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Lecture Comments (6)

0 answers

Post by Kenosha Fox on March 16 at 01:16:57 PM

wow

0 answers

Post by James Whiddon on August 13, 2014

Excellent!

1 answer

Last reply by: Professor Selhorst-Jones
Fri Feb 14, 2014 11:17 AM

Post by DeAnna Dang on February 11, 2014

This helps me so much with my calc class!

1 answer

Last reply by: Professor Selhorst-Jones
Thu Jun 6, 2013 4:25 PM

Post by HAFSA Ahmad on June 6, 2013

wow !this session is excellent.

Instantaneous Slope & Tangents (Derivatives)

  • Long ago in algebra, we were introduced to the concept of slope. We can think of slope as the rate of change that a line has. We define slope as the vertical change divided by the horizontal:
    m  =  rise

    run


     
    .
  • The idea of slope makes sense for a line, because the slope is always the same, no matter where you look. However, on most functions, the slope is constantly changing. While we can't talk about a rate of change/slope for the entire function, we can look for a way to find the instantaneous slope: the slope at a point.
  • Another way to consider the idea of instantaneous slope is through a tangent line: a line going in the "same direction" as a curve at some point, and just touching that one point.
  • We can find the average slope between two points on a function with
    maverage   =   f(x2) − f(x1)

    x2 − x1


     
    .
  • Alternatively, we can rephrase the above formula by considering the average slope between some location x and some location h distance ahead:
    mavg =  f(x+h) − f(x)

    h


     
    .
  • Notice that as h grows smaller and smaller, our slope approximation becomes better and better. With this idea in mind, we take the limit of the above as h→ 0. Assuming the limit exists, we have the derivative:
    f′(x)  =  
    lim
     
      f(x+h) − f(x)

    h


     
    .
    [We read `f′(x)' as `f prime of x'.] By plugging a location into f′(x), we can find the instantaneous slope at that location. We call the process of taking a derivative differentiation.
  • There are many ways to denote the derivative. Given some y = f(x), we can denote it with any of the following:
    f′(x)               dx

    dy


     
                  y′              d

    dx


     


     
    f(x)

     
    The first two are the most common, though.
  • As you progress in calculus, you will quickly learn a wide variety of rules and techniques to make it much easier to find derivatives. You will very seldom, if ever, use the formal definition to find a derivative. The really important idea is what a derivative represents. A derivative is a way to talk about the instantaneous slope (equivalently, rate of change) of a function at some location. No matter how many rules you learn for finding derivatives, never forget that it is, at heart, a way to talk about a function's moment-by-moment change.
  • One rule to make it easier to find the derivative of a function is the power rule. It says that for any function f(x)   =  xn, where n is a constant number, the derivative is
    f′(x)  = n ·xn−1.

Instantaneous Slope & Tangents (Derivatives)

Let f(x) = x2+3x and consider the location x=2. Approximate the slope at that location by using  mavg = [(f(x+h) − f(x))/h] and the following values for h:
h=1,               h=0.1,               h=0.01
  • In the lesson, we learned that
    mavg = f(x+h) − f(x)

    h
    gives the average slope between the x-location and the location that is h ahead of that (in other words, between x and x+h).
  • To find the average slope for any given h-value, we just need to plug in to the formula. We're interested in the location of x=2, so we'll use that for x. For the first h-value, we have h=1, so we set up as below:
    mavg = f(x+h) − f(x)

    h
        ⇒     f(2+1)−f(2)

    1
        =     f(3)−f(2)

    1
    From there, we just use the function to evaluate and simplify:
    f(3)−f(2)

    1
        =     32 + 3(3) − [22 + 3(2)]

    1
        =     9 + 9 −[4+6]     =     8
  • We work the same way for the other h-values. First, h=0.1:
    mavg = f(x+h) − f(x)

    h
        ⇒     f(2+0.1)−f(2)

    0.1
        =     f(2.1)−f(2)

    0.1
    Once h is plugged in, apply the function then evaluate with a calculator:
    f(2.1)−f(2)

    0.1
        =     2.12 + 3(2.1)−[22+3(2)]

    0.1
        =     7.1
    We finish with the last h-value of h=0.01:
    mavg = f(x+h) − f(x)

    h
        ⇒     f(2+0.01)−f(2)

    0.01
        =     f(2.01)−f(2)

    0.01
    Once h is plugged in, apply the function then evaluate with a calculator:
    f(2.01)−f(2)

    0.01
        =     2.012 + 3(2.01)−[22+3(2)]

    0.01
        =     7.01
8,   7.1,   7.01
Let g(x) = √x and consider the location x=4. Approximate the slope at that location by using  mavg = [(g(x+h) − g(x))/h] and the following values for h:
h=1,               h=0.1,               h=0.01
  • In the lesson, we learned that
    mavg = g(x+h) − g(x)

    h
    gives the average slope between the x-location and the location that is h ahead of that (in other words, between x and x+h). [Remark: Notice that the formula was based on f(x) in the lesson and the previous problem. However, the logic behind creating the formula is the same no matter what we call the function, so we can switch over to g(x) without issue.]
  • To find the average slope for any given h-value, we just need to plug in to the formula. We're interested in the location of x=4, so we'll use that for x. For the first h-value, we have h=1, so we set up as below:
    mavg = g(x+h) − g(x)

    h
        ⇒     g(4+1)−g(4)

    1
        =     g(5)−g(4)

    1
    From there, we just apply the function and use a calculator to evaluate:
    g(5)−g(4)

    1
        =     √5− √4

    1
        =     √5 − 2     ≈     0.2361
  • We work the same way for the other h-values. First, h=0.1:
    mavg = g(x+h) − g(x)

    h
        ⇒     g(4+0.1)−g(4)

    0.1
        =     g(4.1)−g(4)

    0.1
    Once h is plugged in, apply the function then evaluate with a calculator:
    g(4.1)−g(4)

    0.1
        =    


     

    4.1
     
    − √4

    0.1
        =    


     

    4.1
     
    − 2

    0.1
        ≈     0.2485
    We finish with the last h-value of h=0.01:
    mavg = g(x+h) − g(x)

    h
        ⇒     g(4+0.01)−g(4)

    0.01
        =     g(4.01)−g(4)

    0.01
    Once h is plugged in, apply the function then evaluate with a calculator:
    g(4.01)−g(4)

    0.01
        =    


     

    4.01
     
    − √4

    0.01
        =    


     

    4.01
     
    − 2

    0.01
        ≈     0.2498
0.2361,     0.2485,     0.2498
Let f(x) = x2 + 3x and consider the location x=2. Find the instantaneous slope at that location by using   limh→ 0 [(f(x+h) − f(x))/h].
  • In the lesson, after learning that the average slope between some x-location and another that is h distance ahead of that is
    mavg = f(x+h) − f(x)

    h
    ,
    we went on to realize that this average slope becomes a more and more accurate approximation of the instantaneous slope at the x-location as h becomes smaller and smaller. While we can't simply plug in h=0 (because it would cause division by 0), we can do the next best thing and consider the limit as h→ 0. Therefore we have that the instantaneous slope at an x-location is given by
    minstant =
    lim
    h→ 0 
    f(x+h) − f(x)

    h
    .
  • Since we're interested in the x-location of x=2, we plug that in:

    lim
    h→ 0 
    f(x+h) − f(x)

    h
        ⇒    
    lim
    h→ 0 
    f(2+h) − f(2)

    h
    Now apply the function for the various inputs:

    lim
    h→ 0 
    f(2+h) − f(2)

    h
        =    
    lim
    h→ 0 
    (2+h)2+3(2+h) − [22 + 3(2)]

    h
  • From here, it's just a matter of evaluating the limit, as we did in previous lessons. [If you don't really know what a limit is, check out the lesson Idea of a Limit. If you're unfamiliar with how to precisely evaluate a limit, check out the lesson Finding Limits.] We can't just plug in h=0 right now, because it would cause division by 0. Instead, we need to simplify and (hopefully) cancel out the h on the bottom that is preventing us from plugging in.

    lim
    h→ 0 
    (2+h)2+3(2+h) − [22 + 3(2)]

    h
        =    
    lim
    h→ 0 
    4+4h + h2+(6+3h) − [4 + 6]

    h
    Continue to simplify, then cancel the common factor of h:

    lim
    h→ 0 
    h2+ 7h +10 − 10

    h
        =    
    lim
    h→ 0 
    h2+ 7h

    h
        =    
    lim
    h→ 0 
    h+ 7
    Now that we've canceled out the h on the bottom of the fraction, we can plug in the limit value without issue:

    lim
    h→ 0 
    h+ 7     =     (0) + 7     =     7
    Thus, at the location of x=2, the instantaneous slope is m=7.

    [Remark: In a previous problem, we worked with f(x) = x2 + 3x and approximated the slope at x=2 by using smaller and smaller values for h. In that problem, as the values got smaller, we got the following slopes:
    8    →     7.1     →     7.01
    Notice that the value these slopes are approaching is 7, exactly what we got when we took the limit as h→ 0. Our previous work matches up with what we did in this problem!]
7
Let g(x) = √x and consider the location x=4. Find the instantaneous slope at that location by using   limh→ 0 [(g(x+h) − g(x))/h]. Once you know the instantaneous slope at that location, use it to find an equation for the tangent line to the curve that passes through the point (4,  g(4)).
  • In the lesson, after learning that the average slope between some x-location and another that is h distance ahead of that is
    mavg = g(x+h) − g(x)

    h
    ,
    we went on to realize that this average slope becomes a more and more accurate approximation of the instantaneous slope at the x-location as h becomes smaller and smaller. While we can't simply plug in h=0 (because it would cause division by 0), we can do the next best thing and consider the limit as h→ 0. Therefore we have that the instantaneous slope at an x-location is given by
    minstant =
    lim
    h→ 0 
    g(x+h) − g(x)

    h
    .
    [Remark: Notice that the formula was based on f(x) in the lesson and the previous problem. However, the logic behind these ideas is the same no matter what we call the function, so we can switch over to g(x) without issue.]
  • Since we're interested in the x-location of x=4, we plug that in:

    lim
    h→ 0 
    g(x+h) − g(x)

    h
        ⇒    
    lim
    h→ 0 
    g(4+h) − g(4)

    h
    Now apply the function for the various inputs:

    lim
    h→ 0 
    g(4+h) − g(4)

    h
        =    
    lim
    h→ 0 


     

    4+h
     
    − √4

    h
        =    
    lim
    h→ 0 


     

    4+h
     
    − 2

    h
  • Now we need to evaluate the limit, as we did in previous lessons. [If you don't really know what a limit is, check out the lesson Idea of a Limit. If you're unfamiliar with how to precisely evaluate a limit, check out the lesson Finding Limits.] We can't just plug in h=0 right now, because it would cause division by 0. Instead, we need to simplify and (hopefully) cancel out the h on the bottom that is preventing us from plugging in. For this problem, we have a square root, so we rationalize the top of the fraction:

    lim
    h→ 0 


     

    4+h
     
    − 2

    h
    ·


     

    4+h
     
    +2



     

    4+h
     
    +2
        =    
    lim
    h→ 0 
    4+h − 4

    h(

     

    4+h
     
    +2)
    From here, we can simplify the top, then cancel out the common factor of h:

    lim
    h→ 0 
    4+h − 4

    h(

     

    4+h
     
    +2)
        =    
    lim
    h→ 0 
    h

    h(

     

    4+h
     
    +2)
        =    
    lim
    h→ 0 
    1



     

    4+h
     
    +2
    Now that we've canceled out the h on the bottom of the fraction, we can plug in the limit value without issue:

    lim
    h→ 0 
    1



     

    4+h
     
    +2
        =     1



     

    4+0
     
    +2
        =     1

    √4+2
        =     1

    2+2
        =     1

    4
    Thus, at the location of x=4, the instantaneous slope is m=[1/4].

    [Remark: In a previous problem, we worked with g(x) = √x and approximated the slope at x=4 by using smaller and smaller values for h. In that problem, as the values got smaller, we got the following slopes:
    0.2361    →     0.2485     →     0.2498
    Notice that the value these slopes are approaching is 0.25, which is the same as what we found when we took the limit as h→ 0. Our previous work matches up with what we did in this problem!]
  • We finish by finding an equation to the tangent line passing through (4,  g(4)). Notice that to find a line equation, you need two things: a point on the line and the slope of the line. From our previous work, we already know the slope at the location we care about: m=[1/4]. That means we just need to work out the point. We're passing through (4,  g(4)), so we'll use that:

    4,  g(4)
        =     (4,  √4)     =     (4,  2)


    Now we have the slope m=[1/4] and the point (4, 2). Let's create our line equation using the slope-intercept form: y=mx+b. We already know m, so we just need to find b. We can solve for b by plugging in the slope and the point:
    y=mx+b     ⇒     2 = 1

    4
    ·4 + b     ⇒     2 = 1+b     ⇒     b=1
    At this point, we know both the slope and the y-intercept (b=1), so we have finished the line equation for the tangent:   y=[(1)/(4)]x + 1.
minstant = [1/4],       Tangent Line:  y=[1/4]x + 1
Let f(x) = x2 + 3x. Find its derivative, f′(x), by using the formal definition of a derivative:
f′(x) =
lim
h→ 0 
  f(x+h)−f(x)

h
Once you find f′(x), use it to find the instantaneous slope at x=−2, x=0, and x=2.
  • From the lesson and the previous problems, we saw that given some specific x-location, we can find the instantaneous slope at the location by plugging in x and taking the limit of the below:
    minstant =
    lim
    h→ 0 
    f(x+h) − f(x)

    h
    .
    The derivative is an extension of that idea. Instead of plugging in a specific x-value, we do the same process but leave x as a variable. This will give us a new function f′(x) that allows us to easily find the instantaneous slope for any x-value. Once we figure out the derivative f′(x), we can simply plug in an x-value to get the slope there.
  • Begin with the definition of the derivative, then apply the function:
    f′(x) =
    lim
    h→ 0 
      f(x+h)−f(x)

    h
        ⇒    
    lim
    h → 0 
    (x+h)2 + 3(x+h) − [x2 + 3x]

    h
    Notice that we can't just plug in h=0 because the expression will "break" from dividing by 0. Instead, we need to simplify the expression first:

    lim
    h → 0 
    x2 + 2xh + h2 + (3x+3h) − [x2 + 3x]

    h
        =    
    lim
    h → 0 
    2xh + h2 + 3h

    h
    Now we can cancel the common factor of h and evaluate the limit:

    lim
    h → 0 
    2x + h + 3     =     2x+(0) +3     =     2x+3
    We have found that the derivative is f′(x) = 2x+3.
  • Now that we have an expression for the derivative, we can easily find the instantaneous slope at any location we want-just plug the x-value in to the derivative:
    f′(−2)    =   2(−2) + 3     =     −4 + 3     =     −1

    f′(0)    =   2(0) + 3     =     0 + 3     =     3

    f′(2)    =   2(2) + 3     =     4 + 3     =     7


    [Remark: In a previous problem, we figured out that for f(x) = x2 + 3x, the instantaneous slope at x=2 comes out to m=7. This is the exact same value that we get when we use the derivative at x=2: f′(2) = 7. Our previous work matches up with what we did in this problem!]
f′(x) = 2x+3,       f′(−2) = −1,     f′(0) = 3,     f′(2) = 7
Let g(x) = √x. Find its derivative, g′(x), by using the formal definition of a derivative:
g′(x) =
lim
h→ 0 
  g(x+h)−g(x)

h
Once you find g′(x), use it to find the instantaneous slope at x=1, x=4, and x=100.
  • From the lesson and the previous problems, we saw that given some specific x-location, we can find the instantaneous slope at the location by plugging in x and taking the limit of the below:
    minstant =
    lim
    h→ 0 
    g(x+h) − g(x)

    h
    .
    The derivative is an extension of that idea. Instead of plugging in a specific x-value, we do the same process but leave x as a variable. This will give us a new function g′(x) that allows us to easily find the instantaneous slope for any x-value. Once we figure out the derivative g′(x), we can simply plug in an x-value to get the slope there. [Remark: Notice that the formulas were based on f(x) in the lesson and the previous problem. However, the logic behind these ideas is the same no matter what we call the function, so we can switch over to g(x) without issue.]
  • Begin with the definition of the derivative, then apply the function:
    g′(x) =
    lim
    h→ 0 
      g(x+h)−g(x)

    h
        ⇒    
    lim
    h → 0 
     


     

    x+h
     
    − √x

    h
    Notice that we can't just plug in h=0 because the expression will "break" from dividing by 0. Instead, we need to simplify the expression first. Since we're dealing with radicals, it will help to rationalize the numerator:

    lim
    h → 0 
     


     

    x+h
     
    − √x

    h
    ·


     

    x+h
     
    + √x



     

    x+h
     
    +√x
        =    
    lim
    h → 0 
      x+h − x

    h(

     

    x+h
     
    +√x)
    Now simplify and cancel the common factor:

    lim
    h → 0 
      h

    h(

     

    x+h
     
    +√x)
        =    
    lim
    h → 0 
      1



     

    x+h
     
    +√x
    At this point the expression will not "break" if we plug in h=0, so we can finish evaluating the limit:

    lim
    h → 0 
      1



     

    x+h
     
    +√x
        =     1



     

    x+0
     
    + √x
        =     1

    √x + √x
        =     1

    2√x
    We have found that the derivative is g′(x) = [1/(2√x)].
  • Now that we have an expression for the derivative, we can easily find the instantaneous slope at any location we want-just plug the x-value in to the derivative:
    g′(1)    =    1

    2√1
        =     1

    2·1
        =     1

    2

    g′(4)    =    1

    2√4
        =     1

    2·2
        =     1

    4

    g′(100)    =    1

    2


    100
        =     1

    2·10
        =     1

    20
    [Remark: In a previous problem, we figured out that for g(x) = √x, the instantaneous slope at x=4 comes out to m=[1/4]. This is the exact same value that we get when we use the derivative at x=4: g′(4) = [1/4]. Our previous work matches up with what we did in this problem!]
g′(x) = [1/(2√x)],       g′(1) = [1/2],    g′(4) = [1/4],    g′(100) = [1/20]
Let P(x) = [1/(x2)]. Find its derivative, P′(x), by using the formal definition of a derivative:
P′(x) =
lim
h→ 0 
  P(x+h)−P(x)

h
Once you know P′(x), use it to find the equation for the tangent line to the curve at (1,  P(1) ).
  • From the lesson and the previous problems, we saw that given some specific x-location, we can find the instantaneous slope at the location by plugging in x and taking the limit of the below:
    minstant =
    lim
    h→ 0 
    P(x+h) − P(x)

    h
    .
    The derivative is an extension of that idea. Instead of plugging in a specific x-value, we do the same process but leave x as a variable. This will give us a new function P′(x) that allows us to easily find the instantaneous slope for any x-value. Once we figure out the derivative P′(x), we can simply plug in an x-value to get the slope there. [Remark: Notice that the formulas were based on f(x) in the lesson and the previous problem. However, the logic behind these ideas is the same no matter what we call the function, so we can switch over to P(x) without issue.]
  • Begin with the definition of the derivative, then apply the function:
    P′(x) =
    lim
    h→ 0 
      P(x+h)−P(x)

    h
        ⇒    
    lim
    h→ 0 
     
    1

    (x+h)2
    1

    x2

    h
    Notice that we can't just plug in h=0 because the expression will "break" from dividing by 0. Instead, we need to simplify the expression first. Since we're dealing with fractions inside a fraction, we want to begin by cleaning that up:

    lim
    h→ 0 
     
    1

    (x+h)2
    1

    x2

    h
    · (x+h)2 (x2)

    (x+h)2 (x2)
        =    
    lim
    h→ 0 
      x2−(x+h)2

    h(x2)(x+h)2
    Simplify and cancel the common factor of h:

    lim
    h→ 0 
      x2−(x2+2xh+h2)

    h(x2)(x+h)2
        =    
    lim
    h→ 0 
      −2xh−h2

    h(x2)(x+h)2
        =    
    lim
    h→ 0 
      −2x−h

    (x2)(x+h)2
    At this point the expression will not "break" if we plug in h=0, so we can finish evaluating the limit:

    lim
    h→ 0 
      −2x−h

    (x2)(x+h)2
        =     −2x−(0)

    (x2)
    x+(0)
    2
     
        =     −2x

    x2(x)2
        =     −2x

    x4
        =     − 2

    x3
    We have found that the derivative is P′(x) = − [2/(x3)].
  • Finally, the problem told us to find an equation for the tangent line that goes through the curve at (1,  P(1) ). To find the equation for a tangent line, we need two things: a point on the line and the slope of the line. We know that the line goes through (1,  P(1) ), so we'll use that point:

    1,  P(1)
        =    
    1,   1

    (1)2

        =     (1,  1)
    Next, we need to find out the slope of the line. Since we want the tangent line to the curve at the above point, we know the line must have the same slope as the instantaneous slope at that location. From the lesson, we learned that the derivative tells us the instantaneous slope at a location, so we use P′(x) to find the slope. We're interested in the x-location of x=1, so plug that in:
    P′(1)    =   − 2

    (1)3
        =     − 2

    1
        =     −2
  • From the above work, we know the tangent line passes through (1,  1) and has a slope of m=−2. Let's use slope-intercept form to find our line equation: y=mx+b. We already know m, so we can plug in our point to solve for b:
    y=mx+b     ⇒     1 = −2(1) + b     ⇒     1 = −2 + b     ⇒     b = 3
    At this point, we know both the slope and the y-intercept (b=3), so we have finished the line equation for the tangent:   y=2x + 3.
P′(x) = − [2/(x3)],        Tangent Line: y=−2x + 3
The power rule is a rule that makes differentiation easier. It says that if we have any function in the format of f(x) = xn, where n is some constant number, then its derivative is just f′(x) = n ·xn−1. Use this rule to find the derivative of f(x) = x6, then use that derivative to find an equation for the tangent line that passes through the point (−1,  f(−1) ).
  • The power rule is shown near the end of the video lesson, between Example 3 and Example 4. We can use it anytime we have a variable raised to the power of some constant number. We take the exponent, put that number in front of the variable as a coefficient, then subtract 1 from the exponent. For example,
    g(x) = x47     ⇒     g′(x) = 47·x46
    [Remark: We've seen this pattern happen in previous problems, even if it was harder to see. Previously, we worked with g(x) = √x and P(x) = [1/(x2)]. Notice that, by the way exponents work, we can rewrite both of those as:
    g(x) = √x     =     x[1/2]              
                  P(x) = 1

    x2
        =     x−2
    Once rewritten in the above format, we see they match the structure needed for the power rule. Applying the power rule to them, we get:
    g(x) = x[1/2]     ⇒     g′(x) = 1

    2
    ·x−[1/2]       
           P(x) = x−2     ⇒     P′(x) = −2 ·x−3
    Finally, by the way exponents work, we can rewrite the derivatives:
    g′(x) = 1

    2
    ·x−[1/2]     =     1

    2√x
                 
                  P′(x) = −2 ·x−3     =     − 2

    x3
    These match perfectly to what we figured out for the derivatives in those previous problems, so this new power rule agrees with our previous work!]
  • The function we're working with is f(x) = x6. Since this is just a variable raised to a constant number, we can apply the power rule. We take the exponent (6), place a copy of that number in front, then lower the exponent by 1:
    f(x) = x6        ⇒        f′(x) = 6 ·x6−1     =     6x5
    Thus the derivative is f(x) = 6x5. [Remark: Notice how much easier it is to use this rule than to apply the formal limit definition of the derivative (where h → 0, etc.). While it's great that this is so easy to do, never lose sight of what the derivative means: the derivative is a way to find the instantaneous slope (rate of change) at some x-location. By plugging in an x-value to f′(x), we find the slope at that place. This is a crucial idea, and it's important to not forget it.]
  • The problem also told us to find an equation for the tangent line that goes through the curve at (−1,  f(−1) ). To find the equation for a tangent line, we need two things: a point on the line and the slope of the line. We know that the line goes through (−1,  f(−1) ), so we'll use that point:

    −1,  f(−1)
        =     (−1,  (−1)6)     =     (−1,  1)
    Next, we need to find out the slope of the line. Since we want the tangent line to the curve at the above point, we know the line must have the same slope as the instantaneous slope at that location. Remember, the derivative tells us the instantaneous slope at a location, so we use f′(x) to find the slope. We're interested in the x-location of x=−1, so plug that in:
    f′(−1)    =   6(−1)5     =     6(−1)     =     −6
  • From the above work, we know the tangent line passes through (−1,  1) and has a slope of m=−6. Let's use slope-intercept form to find our line equation: y=mx+b. We already know m, so we can plug in our point to solve for b:
    y=mx+b     ⇒     1 = −6(−1) + b     ⇒     1 = 6+b     ⇒     b = −5
    At this point, we know both the slope and the y-intercept (b=−5), so we have finished the line equation for the tangent:   y=6x5.
f′(x) = 6x5,        Tangent Line: y=−6x−5
In addition to the power rule, there are a variety of other rules to make differentiation easier. Here are another two. The sum rule says that the derivative of the sum of functions is equal to the sum of those function's derivatives:
Sum Rule:   If f(x) = M(x) + N(x),    then f′(x) = M′(x) + N′(x)
The constant multiple rule says that the derivative of a constant multiplied by a function is equal to the constant multiplied by the function's derivative.
Constant Multiple Rule:    If c is a constant and f(x) = c·M(x) ,    then f′(x) = c ·M′(x)


Let f(x) = 2x2 − 7x +5. Using the above rules and the power rule, find f′(x). Once you have found the derivative f′(x), use it to find an equation for the tangent line that passes through the point (3,  f(3) ).
  • Let's begin by understanding these new rules. First, the sum rule. This says that if a function is made up of multiple parts through addition (or subtraction), we can take the derivative of each part on its own, then just add the parts back together. For example, by the power rule we know that the derivatives to x47 and x3 are 47x46 and 3x2, respectively. That means if we have a function that adds them together (or subtracts one from the other), we can take the derivative as follows:
    g(x) =   x47  + x3        ⇒        g′(x) =   47x46  + 3x2
    Simply put, if two things are separated by just a + or −, we can take the derivative of each thing on its own to find the derivative of the whole.

    Next, the constant multiple rule. This says that if a function is made up of a constant number multiplying something, we can deal with the constant after we take the derivative. For example, by the power rule we know that the derivative of x4 is 4x3. That means if we have a function with a constant number multiplying x4, we can ignore the constant while taking the derivative (although it still multiplies the final result):
    g(x) =   7 x4        ⇒        g′(x) =   7
    4x3
        =     28 x3
    Simply put, if a function is multiplied by a constant, the constant's only effect is to also multiply the derivative.

    [Remark: Notice that the three rules of the power rule, sum rule, and constant multiple rule allow us to take the derivative of any polynomial very easily. This is extremely powerful, since we often wind up working with polynomials. Furthermore, in previous problems we worked with the function f(x) = x2 + 3x. By the power rule, we know that the derivative of x2 is 2x and the derivative of x is 1 (see the next step for an explanation why). Using the sum rule and constant multiple rule with that means we can find the derivative as
    f(x) =   x2 + 3x        ⇒        f′(x) =   2x + 3(1)     =     2x + 3
    This matches perfectly with what we figured out the derivative of x2 +3x was in a previous problem, so these new rules agree with our previous work!]
  • Now that we understand how the rules work, let's work on taking the derivative of f(x) = 2x2 − 7x +5. Since we know how to take the derivative of a variable with a constant exponent, let's rewrite the function so we have that show up in every location (remember, x0=1, so we can put that on the constant without affecting it):
    f(x) = 2x2 − 7x1 +5x0
    Now that we can easily see variables with exponents for every term in the function, let's take the derivative. Remember, the sum rule says we can work with each term on its own (since they're separated by addition/subtraction). Furthermore, the constant multiple rule says we can deal with the constants after we take the derivative. And finally, we'll use the power rule to actually take the derivatives for each variable term:
    f(x) =    
    2
    x2
    7
    x1
    +
    5
    x0)
                
    ⇓          
                
    ⇓          
                
    ⇓          
    f′(x) =    
    2
    2x1
    7
    1x0
    +
    5
    0x−1
    Simplify the result after using the rules:
    f′(x) =   2
    2x1
    − 7
    1x0
    + 5
    0x−1
        =     2(2x) − 7(1) + 5(0)     =     4x − 7
    Thus the derivative is f′(x) = 4x−7. [Remark: It makes sense that the derivatives of −7x and 5 come out as −7 and 0, respectively. Remember, −7x would give a straight line with slope −7. Since the derivative is just a way to find slope and −7x always gives that slope, it makes sense that it comes out be −7. Similarly, 5 would give a horizontal line at height 5. But a horizontal line has a slope of 0, so the derivative of 5 (or any constant by itself, for that matter) comes out to be 0.]
  • The problem also told us to find an equation for the tangent line that goes through the curve at (3,  f(3) ). To find the equation for a tangent line, we need two things: a point on the line and the slope of the line. We know that the line goes through (3,  f(3) ), so we'll use that point:

    3,  f(3)
        =     (3,     2(3)2−7(3)+5)     =     (3,     18−21+5)     =     (3,  2)
    Next, we need to find out the slope of the line. Since we want the tangent line to the curve at the above point, we know the line must have the same slope as the instantaneous slope at that location. Remember, the derivative tells us the instantaneous slope at a location, so we use f′(x) to find the slope. We're interested in the x-location of x=3, so plug that in:
    f′(3)    =   4(3)−7     =     12−7     =     5
  • From the above work, we know the tangent line passes through (3,  2) and has a slope of m=5. Let's use slope-intercept form to find our line equation: y=mx+b. We already know m, so we can plug in our point to solve for b:
    y=mx+b     ⇒     2 = 5(3) + b     ⇒     2 = 15+b     ⇒     b = −13
    At this point, we know both the slope and the y-intercept (b=−13), so we have finished the line equation for the tangent:   y=5x13.
f′(x) = 4x−7,       Tangent Line: y=5x−13
A large water tank currently stands empty. At time t=0 (where t is measured in hours), water begins to flow into the tank from a pipe. The amount of water in the tank can be modeled by the function
V(t) = −58t2 + 1250t,
where the volume V is given in liters.

(a) Using the power rule, sum rule, and constant multiple rule (discussed in previous problems), find the derivative V′(t). (b) Use the derivative you found to give the value of V′(8). (c) Interpret the meaning of your answer to part (b) in the context of the problem: what does the number signify?
  • (a): The differentiation rules work the same no matter what function they are applied to or what variable is involved. [Check out the previous two problems to see what they mean and how they are used.]
    V(t) = −58t2 + 1250t1        ⇒        V′(t) = −58(2t) + 1250(1t0)
    Simplify:
    V′(t) = −58(2t) + 1250(1t0)     =     −116t+1250(1)     =     −116t+1250
  • (b) Now that we know V′(t) = −116t+1250, finding V′(8) is as simple as plugging in t=8:
    V′(8) = −116(8) + 1250     =     −928 +1250     =     322
  • (c) To understand what the value of V′(8) represents, remember that the derivative to a function tells us about its instantaneous slope. However, for this problem, slope doesn't mean much, since we don't precisely care about the shape made by V(t). Instead, remember that talking about slope is the same as talking about how something is changing. With this in mind, we see
    Instantaneous slope     ⇔     Instantaneous rate of change
    Thus, V′(t) is a way to talk about how the original function is changing. To figure out what V′(8) means precisely, remember that V represents the volume of water in the tank, and t=8 is the number of hours after the water begins to flow. Thus V′(8) = 322 is the rate of change at the instant that is 8 hours after the water begins to flow. Finally, since 322 is representing some sort of physical quantity, it needs units. It is the rate of change of volume, so it should involve volume and time. According to the problem, the volume V is measured in liters and the time t is measured in hours, so the rate of change must be given in liters per hour.
(a) V′(t) = −116t+1250 (b) V′(8) = 322 (c) The above represents how the volume of water is changing in the tank. Specifically, it says that 8 hours after the water begins to flow, the volume of water is increasing at a rate of 322 liters per hour (at that moment).

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Instantaneous Slope & Tangents (Derivatives)

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Introduction 0:08
    • The Derivative of a Function Gives Us a Way to Talk About 'How Fast' the Function If Changing
    • Instantaneous Slop
    • Instantaneous Rate of Change
  • Slope 1:24
    • The Vertical Change Divided by the Horizontal
  • Idea of Instantaneous Slope 2:10
    • What If We Wanted to Apply the Idea of Slope to a Non-Line?
  • Tangent to a Circle 3:52
    • What is the Tangent Line for a Circle?
  • Tangent to a Curve 5:20
  • Towards a Derivative - Average Slope 6:36
    • Towards a Derivative - Average Slope, cont.
    • An Approximation
  • Towards a Derivative - General Form 13:18
    • Towards a Derivative - General Form, cont.
    • An h Grows Smaller, Our Slope Approximation Becomes Better
  • Towards a Derivative - Limits! 20:04
    • Towards a Derivative - Limits!, cont.
    • We Want to Show the Slope at x=1
  • Towards a Derivative - Checking Our Slope 23:12
  • Definition of the Derivative 23:54
    • Derivative: A Way to Find the Instantaneous Slope of a Function at Any Point
    • Differentiation
  • Notation for the Derivative 25:58
    • The Derivative is a Very Important Idea In Calculus
  • The Important Idea 27:34
    • Why Did We Learn the Formal Definition to Find a Derivative?
  • Example 1 30:50
  • Example 2 36:06
  • Example 3 40:24
  • The Power Rule 44:16
    • Makes It Easier to Find the Derivative of a Function
    • Examples
    • n Is Any Constant Number
  • Example 4 46:26

Transcription: Instantaneous Slope & Tangents (Derivatives)

Hi--welcome back to Educator.com.0000

Today, we are going to talk about instantaneous slope and tangents, which are also called derivatives.0002

While there are many other things that we could explore in limits, we now have enough of an understanding0008

to move on to another major topic in calculus, the derivative.0012

The derivative of a function gives us a way to talk about how fast the function is changing.0016

It allows us to find the instantaneous slope, which is also called the instantaneous rate of change; they mean equivalent things.0021

And this is a new idea, which we will go over in just a moment.0028

At first glance, knowing a function's rate of change at any location may not seem that useful.0031

But actually, it tells us a massive amount of information.0036

It lets us easily find maximums and minimums; it lets us find increasing or decreasing intervals, and many other things.0039

Knowing the derivative of a function is really, really useful.0046

For example, if we have some function that gives the location of an object,0049

the derivative of that function will tell us the object's velocity, because derivative tells us the rate of change.0054

If we know something's location, and then we talk about the rate of change of that location,0060

well, the rate of change of your location is what your speed is, effectively.0064

That gives us velocity, since velocity is pretty much speed.0068

This is really useful stuff; being able to talk about derivatives of a function is really, really useful, and it forms one of the cornerstones of calculus.0072

Let's check it out: Long, long ago, when we first took algebra, we were introduced to the concept of slope.0079

We can think of slope as the rate of change that a line has, how fast it is moving, in a way.0086

That is, how far up does it go for going some amount horizontally?0093

We define slope as the vertical change, divided by the horizontal, which is also the rise, divided by the run:0098

rise over run, the amount that we have changed vertically, divided by the amount that we have changed horizontally.0108

This tells us how fast our line is changing; it tells us the rate of change, the slope, how much it moves on a moment-by-moment basis for a line.0114

For one step to the right, how much will we go up or down?0123

The idea of slope makes a lot of sense for a line, because its rate of change is always constant.0128

But if we wanted to apply the idea of slope to a non-line, something like, say, a parabola, for example?0133

The first thing to notice is that, for most functions, slope is constantly changing--0139

not necessarily for all functions (for a line, it isn't changing), but for anything that isn't a line, the slope won't be the same everywhere.0144

The rate of change for a function varies depending on what location we consider.0150

For example, on this one, how fast it is changing is totally different in this area; it is totally different from this area and totally different from this area.0154

Each of those three areas is going in a very different rate of change.0162

The way it is moving there is very different.0166

In this area over here, it is mainly going down; in this area over here, it is mainly just going horizontally; and in this area over here, it is mainly going vertically.0168

There is always some horizontal motion in this case, but it ends up changing how it is moving.0178

So, it doesn't have a constant slope; its slope is changing for these things.0183

We want to have some way of being able to talk about what the slope is at this place.0187

What does it change to in this moment?0191

So, we can't talk about a rate of change (slope) for the entire function, but we can look for a way to find the instantaneous slope.0194

What is the slope at some specific point?0201

Here, it is changing at this moment, at some current speed; it is going like this here.0204

But here, it is going like this; or here, it is going like this; and here, it is going like this.0210

We end up getting these different ways of being able to talk about how it is moving in this moment.0219

Where is it going from one spot to the next spot--how is it changing at that place?0224

Another way to work towards this idea of instantaneous slope at a point is through the notion of a tangent line.0229

Now, let's have just a quick break from this: there is a slight relationship between the trigonometric function "tangent" and the "tangent" line of something.0236

But it is really not worth getting into; it is not going to really help us understand things, and the connection is only really tenuous.0245

For now, let's just pretend that they are two totally different ideas, and that they just happen to have the same name.0251

Tangent, when we talk about taking the tangent of some angle, and when we talk about the tangent line on a curve--0258

they are completely unrelated, at least as far as we are concerned right now.0264

It is easier to think about it that way.0268

Back to what we were talking about: for a circle, the tangent line to a point on the circle0270

is a line that passes through the point, but intersects no other part of the circle.0274

Consider this point right here on the circle: the tangent line will go through that point, but it will intersect no other part.0278

We look at it, and we see that we get this right here.0289

See how it barely touches--it just touches, feather-like, that one point on the circle; but it touches nothing else.0293

I want you to notice how the tangent line is basically the instantaneous slope of the curve at that point.0300

If we look in this region, right here, at our point, that is how much the curve is changing at that moment.0306

So, it is as if the tangent line is going the same direction as the curve in that one location.0313

We can take this idea of a tangent line and expand it to any curve.0320

If we have some function f(x), when we draw its graph, we just have a curve on our paper.0323

Now, we can consider some point on the graph and try to find a tangent line at that point.0329

So, say we have some point, like right here, and we want to do the same thing0333

of just barely feather-like touching that one location and going in the same direction as the graph is going.0337

It will pass through that one place, but just barely going in the same direction as the curve is in that moment.0342

We look at that, and we see how that has the instantaneous slope.0350

Notice: the tangent line is the instantaneous slope of the curve at that point.0354

This curve right here has what the slope is at this one place.0359

But if we go to some other place, we would end up having a totally different tangent line for these different points.0365

If a tangent line were to pass through these different points, it would have a totally different slope.0371

The tangent line is going in the same direction as the curve is at that moment, at that single point.0375

At a different point, it might end up having a totally different tangent line, having a totally different slope.0380

So, how can we find this instantaneous slope--what can we do to work towards this slope at some point?0385

Well, let's say we wanted to find the specific instantaneous slope for the function f(x) = x2 + 1 at a horizontal location of x = 1.0392

Here is the point that we are trying to find the instantaneous slope of.0401

And notice that we are currently at the horizontal location x = 1.0404

How could we do this--what could we do to approach it?0409

Well, long ago, when we talked about the properties of functions,0411

we noticed how we could talk about the average slope between two horizontal locations,0414

x1 and x2, on a function, with this formula here.0419

The average slope between these two horizontal locations, x2 and x1, is f(x2) - f(x1)/(x2 - x1.0423

Well, why is that? Well, if we have some other point that is at some x2, what height will that end up being at?0433

Well, that will end up being at a height of...to find the height, we just evaluate the horizontal location.0440

Your input is your horizontal location, and your output is your vertical location.0444

So, that would come out to be f(x2); what input did we have?0448

Well, we put in x1, so we would have an output of f(x1).0451

So, our top, the change, is going to be the difference: f(x2) - f(x1),0456

because we went to f(x2), and we came from f(x1).0466

The difference in our ending and our starting is f(x2) - f(x1).0470

So, that tells us the top part of our average slope formula.0474

The bottom part of it: well, if we go from x1 to x2, then that means that our horizontal motion is going to be x2 - x1.0478

If we go to 10 from 2, we have traveled 8, 10 minus 2; and so, that is why we divide by x2 - x1.0487

It is the rise (how much we have changed vertically), divided by how much we have changed horizontally.0495

So, we have our function, f(x) = x2 + 1; and we want to know what the slope at x = 1 is.0501

What is the value? We have our slope average formula, mavg = f(x2) - f(x1) over (x2 - x1).0506

Now, we want to know the slope at x = 1; so we can use this average slope formula to give us approximations.0515

By using the average slope formula, we can get approximations for what the slope is near x = 1, for finding our instantaneous slope at x = 1.0523

So, since we want to find it near x = 1 (we want to find it specifically at x = 1),0535

let's set the first thing that we will use in our average slope formula as x1 = 1; so we establish this as being x1.0540

For our first approximation, let's use a horizontal location that happens to be two units forward.0549

So, we will have our second horizontal location be moved two forward; we go forward one, forward two; and that makes 3 x2.0554

Or, we could have x2 = x1 + 2; and since we are using x1 = 1, we have 1 + 2, which comes out to be 3.0561

So, we have x1 and x2; we are looking to find the average slope between these two points.0573

All right, our function is f(x) = x2 + 1; we are looking for our instantaneous slope at x = 1.0579

We are working towards that by approximations right now.0586

And our average slope formula is f(x2) - f(x1), over (x2 - x1).0588

We know that x1 = 1, because that is the point that we are interested in finding the slope for.0594

So, we will just set that as sort of a starting place to work from.0599

And we decided, for our first one, to go with x2 = x1 + 2, which was 3.0602

Using our formula, we have mavg = f(x2) - f(x1) over (x2 - x1).0607

x2 is 3; so if we plug that into f(x), then we have f(x2) = 3; so f(x2) is f(3).0613

f(3) would come out to be 32 + 1; 32 + 1 gets us 10.0623

For f(x1), f(x1) would be f(1); f(1) would be 12 + 1, so that gets us 2.0629

So, it is 10 - 2 on the top, and then x2 (3) minus x1 (1), so 3 - 1 on the bottom.0638

We simplify the top; we get 10 - 2, which becomes 8; 3 - 1 becomes 2; 8 divided by 2 gets us 4.0646

And if we draw it in, we end up getting this purple line right here: that is the slope if we set it equal to that average slope.0652

It ends up passing through those two points; so we have an average slope of 4 between those two horizontal locations.0660

OK, that is not a bad start; it is far from perfect--we can clearly see that this line right here is not, in fact, the tangent line.0666

It is not the tangent line; it doesn't pass perfectly against that point; it doesn't just barely, feather-like, touch that one point.0676

It isn't going in the same direction, but it does give us an approximation; it is a start.0682

How can we make this approximation better?0688

Well, we could probably think, "Well, the issue here is the fact that x2 is too far away."0690

We want x2 to be closer; so we can improve it by bringing x2 closer to x1.0694

This time, let's go only one away; we will do x1 + 1, so that it is only one distance forward.0701

So, we will now have x1 + 1, or 2; since we are starting at 1, 1 + 1 gets us 2; now, we are only one horizontal distance away.0706

It is still the same function, x2 + 1; we are still looking for x = 1; it is still the same slope average formula.0716

x1 is still equal to 1, but now x2 is going to be one forward from 1, so that is 2 for our x2.0723

We plug that into our average slope formula; f(x2) is f(2) in this case, because that is our x2 at this place.0730

So, f(x2) would be 22 + 1; 4 + 1 gets us 5;0738

minus f(x1): f(x1) is still 12 + 1, so that still gets us - 2.0743

Our bottom is now x2 - x1; our new x2 is 2, in this case;0749

2 - 1 simplifies to...5 - 2 on top becomes 3; 2 - 1 on the bottom becomes 1.0753

And so, we get a slope of 3; and if we graph that, we end up getting this line right here.0759

All right, nice--we are getting better; it is still not perfect, but the approximation is improving as we bring x2 closer to x1.0766

So, as we bring our x2 closer and closer and closer, we are going to end up getting better and better approximations.0774

What we want to do is bring it really close.0780

Before we can bring x2 really close, though, we need to think about what we are doing in general,0783

so that we can figure out an easy way to formulate talking about bringing x2 really, really close.0787

So, let's talk about what we have been doing, in general.0793

We have our average slope formula; mavg is equal to f(x2) - f(x1), divided by (x2 - x1).0795

That is the output of our second point, minus the output of our first point, divided by the horizontal location difference of our second and first points.0802

So, what we did first was (since we are looking to use this mavg formula): we set x1 at some value.0810

In this case, we set it at the point that we are interested in; we wanted to find the slope of x = 1, so we set our first point,0815

our first horizontal location, as x1 = 1; the first horizontal location is 1, because we want to find out about that slope.0821

Then, from there, we set x2 some distance away from it.0828

The very first time, we set it 2 distance away; so 1 + 2 became 3.0834

The second time we did this, we had 1 + 1 (a horizontal distance of 1); 1 + 1 became 2.0839

So, we want to bring x2 closer and closer by putting less and less distance.0844

What we really care about isn't so much the second point, but the distance to the second point.0848

There are two different ways of looking at this.0853

So, let's call this distance something; we will call it h.0854

What we want to do is bring this h smaller and smaller and smaller.0857

We want to bring x2 closer and closer, so we want to make the distance between our point that we care about,0861

and the point that we are referencing against for our average slope, to become closer and closer and closer.0867

So, if we are calling this distance h, then we can say that x2 is equal to x1, our starting place, plus the distance away, h.0872

So, x2 = x1 + h; with this in mind, we can now rewrite our average slope formula in terms of x1.0880

We started with mavg = f(x2) - f(x1), over (x2 - x1).0888

But now, we have this new way of writing x2: x2 is equal to x1 plus the distance forward, h.0893

So, we can rewrite the formula in terms of x1 and the horizontal distance, h, to be able to create a new formula--0898

not a new formula, so much as a restatement of the old formula--but a new way of looking at it.0906

So, we plug in x2 here; it now becomes x1 + h; so we have x1 + h...0911

f(x1) is still the same; so f(x1 + h) - f(x1); x2 now becomes x1 + h, minus...still x1.0917

On the bottom, we have x1 here and -x1 here; so we can cancel them out.0928

And so, we are left with just h here; on the top, though, we can't cancel anything out,0933

because f(x1 + h) and f(x1)...we don't know how they compare until we apply some specific function to them.0937

So, we are going to have to use the function before we can cancel anything out.0944

We can't cancel out the x1 part inside of the function,0947

because we have to see how h interacts with x1 before we can cancel anything out.0950

OK, one last thing to notice: at this point, the only thing showing up is x1.0955

Only x1 shows up in this formula: we have x1 here, x1 here, x1 here, x1 here, x1 here, x1 here.0962

But no longer x2: we don't have to care about it anymore, because instead we are thinking in terms of this distance h.0969

That is how we swapped to x1 + h.0976

So, if we don't really care about x1 versus x2, then we can just rename x1 as simply x,0978

because at heart, we are all lazy; and it is easy to write out x, compared to x1--just one less thing to write.0985

So, we can now write our average slope as: the average slope is equal to f(x + h) - f(x), divided by h.0992

All right, going back to our thing, we have that our function is x2 + 1 (back to our specific example);1001

and we want to find out what the slope is at a specific value of x = 1.1006

So, for any h, for any distance away from our location x = 1, the average slope is going to be equal to f(x + h) - f(x), all divided by h.1010

Let's see how that is the case: well, if we have x here, and we have x + h here,1024

then the distance forward that we have gone is x + h - x, or simply x.1030

So, that is why we are dividing by h, because we divide by the run; we divide by the horizontal change.1034

And if we want to look at the two heights, well, the height that we will end at will be f(x + h).1038

If we plug x + h in to get an output, it is going to be f acting on (x + h).1043

What is the first one? Our first location is going to be plugging in x, so that would be f(x), f acting on our input of x (so f(x)).1048

What is the distance that we end up having? Well, that will be f(x + h) - f(x).1058

And so, that is why we end up having f(x + h) - f(x) on the top.1062

So, for any distance h that we end up going out, that tells us what the average slope is1067

between our horizontal location x and the h that we end up choosing--whatever distance we end up wanting to use.1073

So, as we make our h smaller and smaller, we will be able to get better and better approximations.1080

We want to see why we end up getting better and better approximations.1084

Notice: let's look at a couple of different points; we choose different points, and we end up running lines through these different slopes.1086

We end up seeing that we get closer and closer to the actual tangent we want.1098

We are getting closer and closer to the tangent we want.1107

So, as we bring our h closer and closer, as we make our h shorter and shorter and shorter, we get better and better approximations.1110

So, as h grows smaller and smaller, our slope approximation becomes better and better.1119

Now, what would be great is if we could somehow set h equal to 0.1124

We want to have the smallest amount of space we can possibly have; the less distance we have, the better our approximation.1129

But if we were to set h equal to 0, then we would be dividing by h.1135

It is f(x + h) - f(x), divided by h; so if h equals 0, we would divide by 0.1138

We can't divide by 0, because that doesn't make sense; it is not defined.1145

So, what we want is...if only there was some way that we could somehow have the same effect as dividing by 0,1149

but not have that issue where we are actually causing it to divide by 0.1155

If only we could look at what it was going to become the instant before it ended up breaking...1158

Limits! That is what that whole thing that we were studying with limits was about!1163

Limits give us a way to talk about what it will become before it breaks--what it is going to become just before it ends up dissolving and not actually making sense.1166

So, we want that infinitesimally small (as h gets really, really, really, really close to 0) thing that ends up happening.1176

What we are looking for is the limit as h goes to 0.1182

As this becomes really, really close to actually being on top of that point, what value do we end up getting out?1185

That is going to be the best sense of what the instantaneous slope is.1192

By getting it really, really, really close, we will be able to get our best idea of what the slope is at that exact place.1196

All right, with this idea in mind, let's take the limit as h goes to 0 of our average slope formula for f(x).1202

So, our average slope formula is mavg = [f(x + h) - f(x)]/h.1209

So, what we do is take the limit as h goes to 0, because the limit as h gets smaller and smaller...1214

our average slope will give us a better and better approximation.1219

As h goes to 0, as h becomes infinitesimally close to it, we will end up getting the best possible approximation.1222

The limit just before the instant it touches--that is the best possible approximation we can get for what the slope is at that place.1228

So, at this point, we can now plug in f(x) = x2 + 1 into our specific f(x) up here1235

and start trying to work towards what the formula is for what the slope will be at that place, at our location x.1241

So, if we have f(x + h), and we are plugging it into f(x) = x2 + 1,1248

we are plugging in x + h into something squared plus 1; so we get (x + h)2 + 1,1254

because that is what our function does; so it is (x + h)2 + 1 for our first portion;1261

and then minus...when we plug in just x, we end up just getting x2 + 1, so - (x2 + 1).1266

And the bottom is just h, because we don't have anything to affect the bottom yet.1273

Limit as h goes to 0...well, we can expand (x + h)2: (x + h)2 becomes x2 + 2xh + h2.1276

The +1 still remains; and -(x2 + 1) is -x2 - 1.1284

At this point, we see that we have positive x2 and negative x2, so they can cancel each other.1291

We have +1 and -1, so they can cancel each other.1296

And we are left with 2xh + h2 on top, all divided by h:1299

the limit as h goes to 0 of 2xh + h2, all divided by h.1304

Great; if we had just plugged in 0 initially, it would end up breaking; we would have 0/0, so we don't end up getting anything out of it.1309

But at this point, we can now cancel things; that is one of the things we talked about when we wanted to evaluate limits.1315

Remember the lesson Finding Limits: how do we find these limits?1319

We get them to a point where we can cancel stuff, so we can see what is going on.1322

So, at this point, we can cancel; we have 2xh + h2, over h, in our limit;1326

so we see that we can cancel this h; that will cancel the h here; and over here, it will cancel the squared and turn it to just h to the 1.1331

So, at this point, we have it simplified to the limit as h goes to 0 of 2x + h.1337

And as h goes to 0, 2x won't be affected; but the h will end up canceling out as it just drops down to 0, and we will be left with 2x.1341

Now, what point did we care about? We cared about the horizontal location.1350

We wanted to show slope at x = 1; so we now have this nice formula to find out what the slope is at some horizontal location.1354

So, we can plug in our x = 1, and we have the instantaneous slope when the horizontal location is x = 1.1363

We know what the slope of f(x) is at the single moment, that single horizontal location, of x = 1.1370

2 times 1...we plug in our value for our x; 2 times 1 comes out to be 2; we have found what the slope is.1376

Let's check--let's see it graphically: if we check this against the graph, we see1386

that a slope of 2 at the horizontal location x = 1 produces a perfect tangent line.1391

This slope of 2 at the horizontal location x = 1, right here, produces a perfect tangent line to the curve at that point.1396

If we draw a line that goes through that point, that has this slope of 2, we end up seeing that it does exactly what we are looking for.1404

It has this bare feather-like touch, just barely on that curve; it just barely touches that one point,1413

and it goes off in the direction that the curve has at that one instant, at that one horizontal location, at that one point; cool.1419

So, this leads us to define the idea of a derivative; we can do what we just did here, for this one specific function, in general.1429

We define the derivative: the derivative is a way to find the instantaneous slope of a function at any point.1435

The derivative of the function f at some horizontal location x is f prime of x (we read this f with this little tick mark as "f prime of x"),1443

and it is the limit as h goes to 0 of f(x + h) - f(x), all divided by h.1454

So, our average slope gives us a better and better sense of what is happening at that instant at that horizontal location x.1462

As our h shrinks down, we get a better and better sense of what is going on from this average slope thing right here.1469

And since we have a better and better sense, the best sense that we will have is the instant before it actually ends up breaking on us at h = 0.1476

So, we take the limit as h goes to 0.1482

Now, of course, this limit has to exist; if the limit doesn't exist, then the derivative doesn't end up working out.1484

But as long as the limit does exist, we manage being able to find out what that instantaneous slope is.1489

We call the process of taking a derivative differentiation; this is called differentiation--we take the derivative through differentiation.1494

You can differentiate a function to get its derivative.1501

And when we write it out, it is just this little tick mark right here, f'(x).1505

You just put this little tick mark right next to your letter that is the letter of the function; and that says the derivative of that function.1511

Once we have the derivative, f'(x), for some function f(x), we can find the instantaneous slope1518

at some specific horizontal location x = a by simply plugging it into our general derivative formula, f'(x).1524

If we want to know the instantaneous slope at some horizontal location a, we just plug it into f'(x), and we have f'(a),1532

just like we did before--we figured out that, in general, for x2 + 1, f prime became 2x;1540

and then we wanted to know what it was at the specific horizontal location of 1.1546

So, we took f'(1); we plugged in 1 for our x, and we got simply 2 as the instantaneous slope at that location.1550

The derivative is a very, very important idea in calculus; it makes one of the absolute cornerstones that calculus is built upon.1558

And so, there end up being a number of different ways to denote it.1565

Given some y = f(x), that is, some function f(x)--or we could talk about it as the vertical location y--1569

we can denote the derivative with any of the following.1577

We can talk about the derivative with any of the following symbols:1579

f'(x), dx/dy, y', d/dx of f of x...and there are even some other ones.1582

In this course, we will end up using f'(x) for the limited period of time that we actually talk about derivatives.1593

But these other ones will end up being used occasionally, as well.1598

And there are reasons why they end up being used; they actually make sense in calculus.1602

We don't quite have time to talk about it right now, but as you work through calculus,1605

as you study calculus, see if you can start to understand why we are talking about it as dx/dy.1608

It has to do with these ideas of infinitesimals; but I will leave that for you, working in your calculus course.1613

All right, the important thing to know is that, while there are all of these different ways to talk about it--1618

we have just 4 right here, but there are even more, occasionally (but you will only end up experiencing really...1622

these two right here are really likely the most common ones you will end up seeing,1627

and this is the only one we will use in this course), they all do the same thing.1631

They represent some function; they represent the derivative that tells us the instantaneous slope for a horizontal location.1636

We plug in a horizontal location, and it tells us what the instantaneous slope is, what the slope is,1643

what the rate of change is at that one specific horizontal location.1648

The important idea of all of this that I really want you to take away (we will work to it) is that,1654

as you progress in calculus, you are quickly going to learn a wide variety of rules.1660

You are going to learn a lot of different rules, a lot of different techniques, that will make it really easier (much easier) to find derivatives.1664

Things that at first seem complicated will actually end up becoming pretty easy, as you learn rules and get used to using rules in calculus.1671

And in actuality, you will very seldom, if ever, use that formal definition that we just saw--1677

that limit as h goes to 0 of f(x + h) minus f(x), over h.1684

That thing won't actually end up getting used a lot; we will mainly end up using these rules and techniques that you will learn as you go through calculus.1688

So, if that is the case--if we end up not really using it that much--why did we learn it? What was the point of learning it?1695

It is because the important part, the reason why we are talking about all of this,1700

is to give you a sense of what the derivative represents before you get to calculus.1703

What we really want to take away from this is that the derivative is a way to talk about instantaneous slope.1707

It is a way to talk about how something is changing, and equivalent to instantaneous slope is the instantaneous rate of change.1714

How is the function changing in this one moment, at this one point, at this one horizontal location?1720

How is it changing--what is the slope right there of the function?1726

It is what it is right there in this function at some specific location; that is the idea of a derivative.1729

So, as you learn these rules, no matter how many rules you learn for finding derivatives,1735

never forget that, at heart, what a derivative is about is a way to talk about a function's moment-by-moment change.1740

It is this idea of how the function is changing right here, right now.1751

What is the slope at this specific place?1755

You will end up learning a lot of rules; you will end up learning a lot of techniques.1758

And it is easy to end up getting tunnel vision and focusing only on the rules and techniques and getting the right numbers out, getting the right symbols out.1761

But even as you are ending up working on this, try to keep that broad idea of what you are thinking about.1767

It is how the thing is changing--how your function is changing, on the whole.1772

You will have to understand how to get those correct values, how to get those correct symbols, when you differentiate--when you take the derivative.1776

But if you forget that the idea of all of this is to talk about the rate of change, you are missing the most important part.1783

The most important part that makes all of this actually have meaning--to be useful--is this idea of how the function is changing here and now.1789

That is why we care about the derivative--not just so that we can churn out numbers;1796

not just so that we can churn out symbols; but so that we can talk about how this thing is changing right here and right now.1800

And by understanding that the derivative represents how this thing is changing right here and right now,1806

and thinking about it in those terms, you will be able to understand all of the larger ideas that we get out of a derivative,1810

of what the derivative represents, and all of the interesting things that a derivative tells us about a function.1816

If you just try to memorize the techniques and rules, and that is all you focus on,1821

you won't have a good sense for what you are doing, and it will become very difficult in calculus.1824

But if you keep this idea in mind of what it represents, it will be easy to understand how things fit together.1828

It will make things a lot more comfortable and make things make a lot more sense.1833

So, the important part of all of this, that I really want you to take away, is to keep the derivative in mind1836

as a way of talking about how the function is changing at some specific location: what is its slope right there?1841

All right, we are ready for some examples.1848

Let f(x) = x3, and consider the location x = -2.1850

We have some function x3, and we are considering the location x = -2.1855

Approximate the slope (that is the instantaneous slope), using our slope average function f(x + h) - f(x) divided by h, and the following values for h.1859

For our first one, h = 2: if our x is at -2, then our x is going to be at -2 for h = 2.1868

x + h, this portion right here, will be equal to...well, if it is -2 for x, and h is 2, then 2 + -2 comes out to be...1877

well, let's write it the other way around: x + h, so -2 + 2.1889

-2 + 2 will come out to be 0; so now, let's plug it into our average slope formula.1894

The average slope is equal to f(x + h) (that was 0), minus f(x) (that is -2, the point we are concerned with),1900

divided by h, the distance we are going out (that is 2).1908

We start working this out; the average slope, f(0)...well, if it is x3, that is 03 - (-2)3, all divided by 2.1911

That comes out to be 0 minus...-2 cubed is -8...over 2; 0 - -8 becomes +8, so we have 8/2, which equals 4.1924

So, this first approximation at h = 2 is: we end up getting an average slope of 4.1937

The next one: h = 1--so once again, our first place will be x = -2, and then we are working from there.1944

x = -2: so x + h will be equal to -2 + 1 (we are going one distance out from -2), which simplifies to -1.1950

So, our average slope between -2 and -1 is going to be f(x + h), so f(-1), minus f(-2), over positive 2.1960

What is f(-1)? Well, that is going to end up being -1 cubed...we already worked out that - f(-2) becomes -(-2)3,1976

which means - -8, which became just +8; so we can just write that as + 8 right now, skipping to that part...divided by 2.1987

-1 cubed becomes -1 times -1 times -1, or just -1, plus 8...1995

Oh, I'm sorry; it is not divided by 2; I'm sorry about that; our h is 1--that is what we are using as our h.1999

I accidentally got stuck on h = 2; we are dividing by 1, because that is the distance out: 1.2005

-1 cubed, plus 8, is -1 + 8; we have 7/1, which equals an average slope of 7 between x = -2 and going a distance of 1 out.2011

Our final one: h = 0.1: our h is still equal to the same starting location, -2, but now we are going out to the very tiny x + h = 2 + 0.1, which gets us 1.9.2024

So, it is just a tiny little bit forward now.2038

Our average slope is going to be f(-2)...sorry, not -2; we do x + h first, and this should be -1.9, because it is -2 + 0.1.2040

So, x + h is f(-1.9), minus f(-2), all divided by our h; that is 0.1.2050

f(-1.9) is going to be -1.9 cubed; we already figured out that minus -2 cubed becomes +8; divide by 0.1.2062

Negative -1.9, cubed, becomes negative 6.859, plus 8, divided by 0.1.2074

So that simplifies, up top, to 1.141 divided by 0.1.2084

We divide by 0.1, and that moves the decimal over, and we get 11.41; great.2091

And so, that gives us our final average slope of h = the tiny distance of 0.1.2100

So, notice: at the very large distance, we got 4; at h = 2, we got 4.2105

At h = 1, we got 7; at h = 0.1, we got 11.41; we are slowly approaching some specific value for what it is.2110

What we are trying to figure out, if we were to draw a graph here, is:2117

here is x3; what is the slope at x = -2?2123

What is the slope that goes through this one place?--that is what we are trying to approximate.2133

What is the m of this tangent line, of that instantaneous slope there?2138

The best that we have gotten so far, when we plugged in this fairly small value of 0.1, was 11.41.2143

If we wanted to get more accurate values from this numerical way of figuring out2148

what the slope starts to work towards, we can use just smaller and smaller h: 0.01, 0.000001...2152

And we will be able to get better and better, more accurate, results.2158

However, if we want to get it perfectly, we have to use the derivative--a problem about the derivative!--2161

for f(x) = x3, then evaluating f'(-2).2165

And so, once again, f'(-2) tells us the instantaneous slope of our graph at this value, -2.2169

So, figuring out f(-2) will tell us the slope; the slope of this is going to be this value that we get from f'(-2).2179

All right, so how do we figure out f prime?2189

Well, remember: f'(x) is equal to the limit as h goes to 0 of f(x + h) minus f(x), all over h.2191

In this case, our f(x) is equal to x3, so we can swap this out for the limit as h goes to 0 of f(x + h)...2205

so f(x + h) becomes (x + h)3--we are plugging in (x + h) into how it works over here,2219

so (x + h)3, minus...now we are just plugging in x, so simply x3, all over h.2228

What is (x + h)3? Let's just work that out in a quick sidebar.2235

(x + h)3: we can write that as (x + h)(x + h)2; that becomes x2 + 2xh + h2...2239

(x + h) times x2 + 2xh + h2...we get x3;2253

x times 2xh is 2x2h; h times x2 is 1hx2, or 1x2h.2258

So, we get + 3x2h; then, x times h2 is 1xh2; h times 2xh is 2xh2;2263

so we get a total of 3xh2, plus h3; h2 times h gets us h3.2274

Notice that we can also get that through the binomial expansion, if you remember.2280

If you recently worked on the binomial expansion, you might recognize that.2285

All right, so at that point, we can plug back into our limit, limit as h goes to 0...2288

we swap out (x + h)3 for x3 + 3x2h + 3xh2 + h3 - x3, all divided by h.2293

At this point, we see that we have a positive x3 and a negative x3; they knock each other out.2308

Now, we see that everything has an h; great.2313

Currently, if we were to just plug in h goes to 0, we would get 0 + 0 + 0, divided by 0; 0/0...we can't do that.2316

But we can knock out the dividing by 0, because everything up top now has a factor of h in it.2324

So, we can knock out this h here; that knocks the h out here, turns this h2 into an h1,2329

turns this h3 into an h2...and we now have the limit as h goes to 0 of 3x2 + 3xh + h2.2334

The limit as h goes to 0...well, that is going to cause 3xh to just turn into 0.2349

As h goes to 0, it will crush the x; as h2 goes to 0, it will crush h2 to 0.2355

So, as h goes to 0, any h2 will crush h2 to 0; and that leaves us with just 3x2; great.2361

So, f'(x) is equal to 3x2.2368

So now, we were asked to find what is the specific derivative at f'(-2).2374

f'(-2) = what? Well, f'(x) = 3x2, so f'(-2) will be 3(-2)2.2379

3 times -2 squared...that becomes 4, so we have 3 times 4; and we get 12.2391

The instantaneous slope of the point, of the horizontal location -2, is a slope of 12,2398

which, if you can remember from that last example that we were just working to... when we had .1, we managed to get to 11.41.2404

As we were making our h smaller and smaller, we were slowly working our way2411

to this perfectly instantaneous slope of 12, working our way to the derivative at -2; cool.2414

The next one: Find the derivative of f(x) = 1/x.2421

The same basic method works here: limit as h goes to 0 of...I'm sorry, f prime, the derivative...2425

f'(x) is going to be equal to the limit as h goes to of f(x + h) - f(x), all over h.2433

So, we start working this out: limit as h goes to 0 of f(x + h) - f(x)...2442

Well, x + h...we will plug into our formula f(x) = 1/x, so f(thing) = 1/thing.2446

So, if we plug in x + h for our thing, we have 1/(x + h), minus f(x) (is simply going to be 1/x), all divided by h.2453

Oh, I forgot one important part: it equals the limit as h goes to 0 of that stuff.2463

We have to keep up that limit, because it is very important.2468

Limit as h goes to 0 of all of this stuff: now, from when we worked with finding limits,2471

well, we have fractions on top of and inside of a fraction...2476

Well, what we want is less fractions; we don't want fractions in fractions, so how can we get rid of the fractions up top?2479

Well, notice: if we multiply by the top, we can cancel out those fractions with x times x + h.2484

That will cancel out the right fraction and the left fraction on the top.2490

We will be left with some stuff...but we have to multiply the top and the bottom of any fraction by the same thing; otherwise, we just have wishful thinking.2494

x times (x + h) on the top and the bottom: we have...this is equal to the limit as h goes to 0 of 1/(x + h) times (x + h)...2501

well, that (x + h) here will cancel out the (x + h) here, and we will be left with just x.2511

We get x, minus quantity...here, the x cancels out the x here, and we are going to be left with x + h, minus (x + h), all over...2517

well, we can't cancel out anything on the bottom; we just have h.2528

And if you remember from our finding limits, it is actually going to behoove us to be able to keep the h there,2531

because our goal, since we are going to have h go to 0...we can't divide by 0;2535

so we need to somehow get that h at the bottom to cancel out and be canceled out.2539

Otherwise, our limit will end up getting mixed up by that h still being there.2543

So, h times x times x + h is on the bottom.2547

At this point, we see that we have x - (x + h); well, put -h in here...2550

Oh, I accidentally cut out the wrong thing there; I should not have crossed out the h; I should have crossed out the x.2557

The -x here cancels out the positive x here; and we are left with the limit as h goes to 0 of -h, divided by h times x times (x + h); great.2567

So, we have the h here on the bottom and the h here on the top.2584

So, we can cancel out h here and h here; that leaves us with -1 on the top:2589

the limit as h goes to 0 of -1, all divided by x times x + h.2594

At this point, if h goes to 0, we won't have massive issues.2602

Let's see one more: the limit as h goes to 0 of -1 over...we expand x and (x + h); we distribute that x over, and we get x2 + xh.2606

Now, as h goes to 0, that will cause the xh to cancel out; but it will have no effect on the x2,2620

so things don't end up breaking, as long as x isn't equal to 0.2626

But we are looking just in general; so -1/x2 is what ends up coming out of this: -1/x2.2629

So, that means that we can write, in general, f'(x) is equal to what we ended up getting of all of this in the end, -1/x2; great.2636

All right, finally, before we get to our final, fourth example, let's talk about the power rules.2650

These are the sorts of rules that you will end up learning as you start working through calculus.2655

One of the first and easiest rules to learn, that makes it so much easier to take a derivative, is the power rule.2659

What it says is that, for any function f(x) = xn, where n is just any constant number, the derivative of f(x) is f'(x) = n(xn - 1).2665

So, we take that power; we bring it down in front of it; we have x to the n, and then we bring it down in front;2680

we have n times...and then we bring down our n to n - 1.2690

So, you might not have noticed this yet, but this actually ended up holding true2695

for all of the functions that we have worked through so far in this lesson, both in the early part of the lesson,2699

where we were setting up these ideas, and in Example 2 and Example 3.2703

When we took the derivative of f(x) = x2 + 1, it came out to be simply 2x.2707

Don't be too worried about the +1; we can think of it is being x0.2712

So, when you bring down that 0 in front of it, it just cancels out and becomes nothing.2716

So, we got 2x out of it.2719

When we had f(x) = x3, when we took the derivative of that in Example 2,2721

we ended up getting 3 times x2: the 3 went down in front, and we brought down 3 by 1 to 2; 3 - 1 becomes 2.2726

And finally, Example 3, that we just worked on: f(x) = x-1...we brought down the -1, and we got -x; and -1 - 1 becomes -2.2736

So, we ended up seeing this inadvertently.2746

And this ends up being true for any constant number n; it really works for anything at all.2749

If you are curious about seeing this some more, try looking at what would happen if you tried to take the derivative2755

through that formal h goes to 0 of x4, f(x4); try working through the formal definition of x4.2758

And if you really want to try seeing how it works for any integer at all, or any positive integer, at least, try using the binomial theorem.2765

And it even works for anything at all, as we have just seen for negative numbers.2773

But you can end up seeing how it works for binomial theorem, as well.2776

If you are curious about this, try checking it out; it is pretty easy to end up seeing,2779

with the binomial theorem, that it ends up being true for any positive integer.2782

All right, let's put this to use: using that power rule, find an equation for the tangent line to the function f(x) = x4 that passes through (3,81).2785

If we are going to pass through (3,81), let's first get a sense of what is going on.2797

Let's draw just a really quick sketch of what is going on here.2800

x4 shoots up really, really quickly; we shoot up massively, very, very quickly, with x4.2803

By the time we have made it to a horizontal x location of 3, we are putting out an output of 81; 34 is 9 times 9, which is 81.2810

So, we are at this very, very high point, very, very quickly.2818

So, what we are going to want to find is: we want to find the tangent line to the function at this point, (3,81).2822

We want to find something that ends up going like this; that is what we are looking for: what is the slope at this?2829

Well, to find the slope at any given horizontal location, at any point, we end up taking the derivative.2835

How can we take the derivative? If we have f(x) = x4, the power rule says that we can get the derivative,2842

f'(x), by taking the exponent and bringing it down in the front; so we have 4 times x;2852

and the exponent goes...subtract by 1; so 4 - 1 becomes 4x3.2861

So, if we want to find out what the slope is at x = 3 (that is the horizontal location we are considering),2867

if we want to find out what the slope is for our tangent line, the slope at x = 3 is going to be f'(3).2879

f'(3) will be 4 times 33; we plug it into our f prime, our derivative function.2885

f'(3) = 4(3)3; 3 cubed is 27; 4 times 27 equals 108.2893

So, we now know that the slope of our tangent line is 108.2902

So, if we are going to work out the tangent line: our tangent line is going to end up using this slope, m = 108.2907

So, any line at all can be described by slope-intercept form; it is a really good form to have memorized: y = mx + b.2917

You always want to keep that one handy; it is always useful.2925

y = mx + b: well, we just figured out that the slope...f it is going to be the tangent, it must have a slope of 108, because f prime,2928

the instantaneous slope at 3, f'(3), comes out to be 108.2938

So, we know that the instantaneous slope at the point we are interested in, (3,81), is 108.2942

So, that means that we have y = 108x + some b that we haven't figured out yet.2948

How can we figure out what b is to finish creating our equation for the tangent line?2957

If you are going to figure out what any line is, you need to know2962

what the slope is and what the y-intercept is--what m is and what b is, at least if you are using slope-intercept form.2966

y = 108x + b; well, do we know any points on that line?2972

Yes, we are looking at the tangent line; and we were told that the tangent line passes through the point (3,81).2977

So, we can plug in the point (3,81), because we know that our tangent line2984

has to pass through the one point that it barely, barely just touches on that curve.2989

So, we plug in (3,81); that is 81 for our y-value; that equals 108 times 3 (for our horizontal value x) plus b.2993

81 = 108 times 3 (that comes out to be 324), plus b; 81 - 324 is -243 for our b.3003

So now, we know that -243 = b; we know what our slope is; so that means that we can describe the tangent line in general as y = 108x - 243.3014

That is y = mx + b with our m and b filled in.3027

And now, we have the tangent line that passes through the point (3,81) and is tangent to the curve created by x4--pretty cool.3031

Calculus gives us a whole bunch of stuff that we can end up doing with the derivative.3039

The derivative is this massively, massively useful, important thing.3042

And we are just barely touching the surface of how incredibly useful and cool this thing is.3045

As you work through calculus, you will end up learning a whole bunch of things that the derivative lets us do.3050

It lets us learn about a function; it is really, really amazing how much information it gives us; calculus is really, really cool.3054

I hope that, at some point, you get the chance to take calculus and get to see how many cool things there are.3060

And remember: when you work with the derivative, what you are looking at is what the slope is of that location,3064

of that point, of that horizontal location on your graph.3068

All right, we will see you at Educator.com later--goodbye!3071