For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
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
m_{average} = f(x_{2}) − f(x_{1}) x_{2} − x_{1}
.  Alternatively, we can rephrase the above formula by considering the average slope between some location x and some location h distance ahead:
m_{avg} = 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:
[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.f′(x) =
lim
f(x+h) − f(x) h
.  There are many ways to denote the derivative. Given some y = f(x), we can denote it with any of the following:
The first two are the most common, though.f′(x) dx dy
y′ d dx
⎡
⎣
f(x) ⎤
⎦
 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 momentbymoment 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) = x^{n}, where n is a constant number, the derivative is
f′(x) = n ·x^{n−1}.
Instantaneous Slope & Tangents (Derivatives)

 In the lesson, we learned that
gives the average slope between the xlocation and the location that is h ahead of that (in other words, between x and x+h).m_{avg} = f(x+h) − f(x) h  To find the average slope for any given hvalue, 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 hvalue, we have h=1, so we set up as below:
From there, we just use the function to evaluate and simplify:m_{avg} = f(x+h) − f(x) h⇒ f(2+1)−f(2) 1= f(3)−f(2) 1f(3)−f(2) 1= 3^{2} + 3(3) − [2^{2} + 3(2)] 1= 9 + 9 −[4+6] = 8  We work the same way for the other hvalues. First, h=0.1:
Once h is plugged in, apply the function then evaluate with a calculator:m_{avg} = f(x+h) − f(x) h⇒ f(2+0.1)−f(2) 0.1= f(2.1)−f(2) 0.1
We finish with the last hvalue of h=0.01:f(2.1)−f(2) 0.1= 2.1^{2} + 3(2.1)−[2^{2}+3(2)] 0.1= 7.1
Once h is plugged in, apply the function then evaluate with a calculator:m_{avg} = f(x+h) − f(x) h⇒ f(2+0.01)−f(2) 0.01= f(2.01)−f(2) 0.01f(2.01)−f(2) 0.01= 2.01^{2} + 3(2.01)−[2^{2}+3(2)] 0.01= 7.01

 In the lesson, we learned that
gives the average slope between the xlocation 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.]m_{avg} = g(x+h) − g(x) h  To find the average slope for any given hvalue, 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 hvalue, we have h=1, so we set up as below:
From there, we just apply the function and use a calculator to evaluate:m_{avg} = g(x+h) − g(x) h⇒ g(4+1)−g(4) 1= g(5)−g(4) 1g(5)−g(4) 1= √5− √4 1= √5 − 2 ≈ 0.2361  We work the same way for the other hvalues. First, h=0.1:
Once h is plugged in, apply the function then evaluate with a calculator:m_{avg} = g(x+h) − g(x) h⇒ g(4+0.1)−g(4) 0.1= g(4.1)−g(4) 0.1
We finish with the last hvalue of h=0.01:g(4.1)−g(4) 0.1=
√4.1− √4 0.1=
√4.1− 2 0.1≈ 0.2485
Once h is plugged in, apply the function then evaluate with a calculator:m_{avg} = g(x+h) − g(x) h⇒ g(4+0.01)−g(4) 0.01= g(4.01)−g(4) 0.01g(4.01)−g(4) 0.01=
√4.01− √4 0.01=
√4.01− 2 0.01≈ 0.2498
 In the lesson, after learning that the average slope between some xlocation and another that is h distance ahead of that is
we went on to realize that this average slope becomes a more and more accurate approximation of the instantaneous slope at the xlocation 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 xlocation is given bym_{avg} = f(x+h) − f(x) h, m_{instant} =
lim
h→ 0f(x+h) − f(x) h.  Since we're interested in the xlocation of x=2, we plug that in:
Now apply the function for the various inputs:
lim
h→ 0f(x+h) − f(x) h⇒
lim
h→ 0f(2+h) − f(2) h
lim
h→ 0f(2+h) − f(2) h=
lim
h→ 0(2+h)^{2}+3(2+h) − [2^{2} + 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.
Continue to simplify, then cancel the common factor of h:
lim
h→ 0(2+h)^{2}+3(2+h) − [2^{2} + 3(2)] h=
lim
h→ 04+4h + h^{2}+(6+3h) − [4 + 6] h
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→ 0h^{2}+ 7h +10 − 10 h=
lim
h→ 0h^{2}+ 7h h=
lim
h→ 0h+ 7
Thus, at the location of x=2, the instantaneous slope is m=7.
lim
h→ 0h+ 7 = (0) + 7 = 7
[Remark: In a previous problem, we worked with f(x) = x^{2} + 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:
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!]8 → 7.1 → 7.01
 In the lesson, after learning that the average slope between some xlocation and another that is h distance ahead of that is
we went on to realize that this average slope becomes a more and more accurate approximation of the instantaneous slope at the xlocation 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 xlocation is given bym_{avg} = 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.]m_{instant} =
lim
h→ 0g(x+h) − g(x) h.  Since we're interested in the xlocation of x=4, we plug that in:
Now apply the function for the various inputs:
lim
h→ 0g(x+h) − g(x) h⇒
lim
h→ 0g(4+h) − g(4) h
lim
h→ 0g(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:
From here, we can simplify the top, then cancel out the common factor of h:
lim
h→ 0
√4+h− 2 h·
√4+h+2
√4+h+2 =
lim
h→ 04+h − 4 h(
√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→ 04+h − 4 h(
√4+h+2) =
lim
h→ 0h h(
√4+h+2) =
lim
h→ 01
√4+h+2
Thus, at the location of x=4, the instantaneous slope is m=[1/4].
lim
h→ 01
√4+h+2 = 1
√4+0+2 = 1 √4+2= 1 2+2= 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:
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!]0.2361 → 0.2485 → 0.2498  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 slopeintercept 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:
At this point, we know both the slope and the yintercept (b=1), so we have finished the line equation for the tangent: y=[(1)/(4)]x + 1.y=mx+b ⇒ 2 = 1 4·4 + b ⇒ 2 = 1+b ⇒ b=1

 From the lesson and the previous problems, we saw that given some specific xlocation, we can find the instantaneous slope at the location by plugging in x and taking the limit of the below:
The derivative is an extension of that idea. Instead of plugging in a specific xvalue, 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 xvalue. Once we figure out the derivative f′(x), we can simply plug in an xvalue to get the slope there.m_{instant} =
lim
h→ 0f(x+h) − f(x) h.  Begin with the definition of the derivative, then apply the function:
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:f′(x) =
lim
h→ 0f(x+h)−f(x) h⇒
lim
h → 0(x+h)^{2} + 3(x+h) − [x^{2} + 3x] h
Now we can cancel the common factor of h and evaluate the limit:
lim
h → 0x^{2} + 2xh + h^{2} + (3x+3h) − [x^{2} + 3x] h=
lim
h → 02xh + h^{2} + 3h h
We have found that the derivative is f′(x) = 2x+3.
lim
h → 02x + h + 3 = 2x+(0) +3 = 2x+3  Now that we have an expression for the derivative, we can easily find the instantaneous slope at any location we wantjust plug the xvalue 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) = x^{2} + 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!]

 From the lesson and the previous problems, we saw that given some specific xlocation, we can find the instantaneous slope at the location by plugging in x and taking the limit of the below:
The derivative is an extension of that idea. Instead of plugging in a specific xvalue, 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 xvalue. Once we figure out the derivative g′(x), we can simply plug in an xvalue 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.]m_{instant} =
lim
h→ 0g(x+h) − g(x) h.  Begin with the definition of the derivative, then apply the function:
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:g′(x) =
lim
h→ 0g(x+h)−g(x) h⇒
lim
h → 0
√x+h− √x h
Now simplify and cancel the common factor:
lim
h → 0
√x+h− √x h·
√x+h+ √x
√x+h+√x =
lim
h → 0x+h − x h(
√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 → 0h h(
√x+h+√x) =
lim
h → 01
√x+h+√x
We have found that the derivative is g′(x) = [1/(2√x)].
lim
h → 01
√x+h+√x = 1
√x+0+ √x = 1 √x + √x= 1 2√x  Now that we have an expression for the derivative, we can easily find the instantaneous slope at any location we wantjust plug the xvalue in to the derivative:
g′(1) = 1 2√1= 1 2·1= 1 2g′(4) = 1 2√4= 1 2·2= 1 4
[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′(100) = 1 2
√100= 1 2·10= 1 20

 From the lesson and the previous problems, we saw that given some specific xlocation, we can find the instantaneous slope at the location by plugging in x and taking the limit of the below:
The derivative is an extension of that idea. Instead of plugging in a specific xvalue, 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 xvalue. Once we figure out the derivative P′(x), we can simply plug in an xvalue 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.]m_{instant} =
lim
h→ 0P(x+h) − P(x) h.  Begin with the definition of the derivative, then apply the function:
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:P′(x) =
lim
h→ 0P(x+h)−P(x) h⇒
lim
h→ 01 (x+h)^{2}− 1 x^{2}h
Simplify and cancel the common factor of h:
lim
h→ 01 (x+h)^{2}− 1 x^{2}h· (x+h)^{2} (x^{2}) (x+h)^{2} (x^{2})=
lim
h→ 0x^{2}−(x+h)^{2} h(x^{2})(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→ 0x^{2}−(x^{2}+2xh+h^{2}) h(x^{2})(x+h)^{2}=
lim
h→ 0−2xh−h^{2} h(x^{2})(x+h)^{2}=
lim
h→ 0−2x−h (x^{2})(x+h)^{2}
We have found that the derivative is P′(x) = − [2/(x^{3})].
lim
h→ 0−2x−h (x^{2})(x+h)^{2}= −2x−(0) (x^{2}) ⎛
⎝x+(0) ⎞
⎠2
= −2x x^{2}(x)^{2}= −2x x^{4}= − 2 x^{3}  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:
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 xlocation of x=1, so plug that in:⎛
⎝1, P(1) ⎞
⎠= ⎛
⎝1, 1 (1)^{2}⎞
⎠= (1, 1) 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 slopeintercept form to find our line equation: y=mx+b. We already know m, so we can plug in our point to solve for b:
At this point, we know both the slope and the yintercept (b=3), so we have finished the line equation for the tangent: y=−2x + 3.y=mx+b ⇒ 1 = −2(1) + b ⇒ 1 = −2 + b ⇒ b = 3
 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,
[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/(x^{2})]. Notice that, by the way exponents work, we can rewrite both of those as:g(x) = x^{47} ⇒ g′(x) = 47·x^{46}
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 = x^{[1/2]} ⎢
⎢P(x) = 1 x^{2}= x^{−2}
Finally, by the way exponents work, we can rewrite the derivatives:g(x) = x^{[1/2]} ⇒ g′(x) = 1 2·x^{−[1/2]} ⎢
⎢P(x) = x^{−2} ⇒ P′(x) = −2 ·x^{−3}
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!]g′(x) = 1 2·x^{−[1/2]} = 1 2√x⎢
⎢P′(x) = −2 ·x^{−3} = − 2 x^{3}  The function we're working with is f(x) = x^{6}. 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:
Thus the derivative is f(x) = 6x^{5}. [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 xlocation. By plugging in an xvalue to f′(x), we find the slope at that place. This is a crucial idea, and it's important to not forget it.]f(x) = x^{6} ⇒ f′(x) = 6 ·x^{6−1} = 6x^{5}  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:
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 xlocation of x=−1, so plug that in:⎛
⎝−1, f(−1) ⎞
⎠= (−1, (−1)^{6}) = (−1, 1) 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 slopeintercept form to find our line equation: y=mx+b. We already know m, so we can plug in our point to solve for b:
At this point, we know both the slope and the yintercept (b=−5), so we have finished the line equation for the tangent: y=−6x −5.y=mx+b ⇒ 1 = −6(−1) + b ⇒ 1 = 6+b ⇒ b = −5


Let f(x) = 2x^{2} − 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 x^{47} and x^{3} are 47x^{46} and 3x^{2}, 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:
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.g(x) = x^{47} + x^{3} ⇒ g′(x) = 47x^{46} + 3x^{2}
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 x^{4} is 4x^{3}. That means if we have a function with a constant number multiplying x^{4}, we can ignore the constant while taking the derivative (although it still multiplies the final result):
Simply put, if a function is multiplied by a constant, the constant's only effect is to also multiply the derivative.g(x) = 7 x^{4} ⇒ g′(x) = 7 ⎛
⎝4x^{3} ⎞
⎠= 28 x^{3}
[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) = x^{2} + 3x. By the power rule, we know that the derivative of x^{2} 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
This matches perfectly with what we figured out the derivative of x^{2} +3x was in a previous problem, so these new rules agree with our previous work!]f(x) = x^{2} + 3x ⇒ f′(x) = 2x + 3(1) = 2x + 3  Now that we understand how the rules work, let's work on taking the derivative of f(x) = 2x^{2} − 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, x^{0}=1, so we can put that on the constant without affecting it):
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) = 2x^{2} − 7x^{1} +5x^{0}
Simplify the result after using the rules:f(x) = 2 ⎛
⎝x^{2} ⎞
⎠− 7 ⎛
⎝x^{1} ⎞
⎠+ 5 ⎛
⎝x^{0}) ⇓ ⇓ ⇓ f′(x) = 2 ⎛
⎝2x^{1} ⎞
⎠− 7 ⎛
⎝1x^{0} ⎞
⎠+ 5 ⎛
⎝0x^{−1} ⎞
⎠
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.]f′(x) = 2 ⎛
⎝2x^{1} ⎞
⎠− 7 ⎛
⎝1x^{0} ⎞
⎠+ 5 ⎛
⎝0x^{−1} ⎞
⎠= 2(2x) − 7(1) + 5(0) = 4x − 7  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:
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 xlocation of x=3, so plug that in:⎛
⎝3, f(3) ⎞
⎠= (3, 2(3)^{2}−7(3)+5) = (3, 18−21+5) = (3, 2) 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 slopeintercept form to find our line equation: y=mx+b. We already know m, so we can plug in our point to solve for b:
At this point, we know both the slope and the yintercept (b=−13), so we have finished the line equation for the tangent: y=5x −13.y=mx+b ⇒ 2 = 5(3) + b ⇒ 2 = 15+b ⇒ b = −13

(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.]
Simplify:V(t) = −58t^{2} + 1250t^{1} ⇒ V′(t) = −58(2t) + 1250(1t^{0}) V′(t) = −58(2t) + 1250(1t^{0}) = −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
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.Instantaneous slope ⇔ Instantaneous rate of change
*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 screencaptured 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
 Introduction
 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
 Idea of Instantaneous Slope
 Tangent to a Circle
 Tangent to a Curve
 Towards a Derivative  Average Slope
 Towards a Derivative  General Form
 Towards a Derivative  General Form, cont.
 An h Grows Smaller, Our Slope Approximation Becomes Better
 Towards a Derivative  Limits!
 Towards a Derivative  Checking Our Slope
 Definition of the Derivative
 Notation for the Derivative
 The Important Idea
 Example 1
 Example 2
 Example 3
 The Power Rule
 Example 4
 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 NonLine?
 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
Precalculus with Limits Online Course
Transcription: Instantaneous Slope & Tangents (Derivatives)
Hiwelcome 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 momentbymoment 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 nonline, something like, say, a parabola, for example?0133
The first thing to notice is that, for most functions, slope is constantly changing0139
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 spothow 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 curve0258
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 touchesit just touches, featherlike, 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 featherlike 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 slopewhat 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) = x^{2} + 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 thiswhat 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
x_{1} and x_{2}, on a function, with this formula here.0419
The average slope between these two horizontal locations, x_{2} and x_{1}, is f(x_{2})  f(x_{1})/(x_{2}  x_{1}.0423
Well, why is that? Well, if we have some other point that is at some x_{2}, 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(x_{2}); what input did we have?0448
Well, we put in x_{1}, so we would have an output of f(x_{1}).0451
So, our top, the change, is going to be the difference: f(x_{2})  f(x_{1}),0456
because we went to f(x_{2}), and we came from f(x_{1}).0466
The difference in our ending and our starting is f(x_{2})  f(x_{1}).0470
So, that tells us the top part of our average slope formula.0474
The bottom part of it: well, if we go from x_{1} to x_{2}, then that means that our horizontal motion is going to be x_{2}  x_{1}.0478
If we go to 10 from 2, we have traveled 8, 10 minus 2; and so, that is why we divide by x_{2}  x_{1}.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) = x^{2} + 1; and we want to know what the slope at x = 1 is.0501
What is the value? We have our slope average formula, m_{avg} = f(x_{2})  f(x_{1}) over (x_{2}  x_{1}).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 x_{1} = 1; so we establish this as being x_{1}.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 x_{2}.0554
Or, we could have x_{2} = x_{1} + 2; and since we are using x_{1} = 1, we have 1 + 2, which comes out to be 3.0561
So, we have x_{1} and x_{2}; we are looking to find the average slope between these two points.0573
All right, our function is f(x) = x^{2} + 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(x_{2})  f(x_{1}), over (x_{2}  x_{1}).0588
We know that x_{1} = 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 x_{2} = x_{1} + 2, which was 3.0602
Using our formula, we have m_{avg} = f(x_{2})  f(x_{1}) over (x_{2}  x_{1}).0607
x_{2} is 3; so if we plug that into f(x), then we have f(x_{2}) = 3; so f(x_{2}) is f(3).0613
f(3) would come out to be 3^{2} + 1; 3^{2} + 1 gets us 10.0623
For f(x_{1}), f(x_{1}) would be f(1); f(1) would be 1^{2} + 1, so that gets us 2.0629
So, it is 10  2 on the top, and then x_{2} (3) minus x_{1} (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 perfectwe 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, featherlike, 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 x_{2} is too far away."0690
We want x_{2} to be closer; so we can improve it by bringing x_{2} closer to x_{1}.0694
This time, let's go only one away; we will do x_{1} + 1, so that it is only one distance forward.0701
So, we will now have x_{1} + 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, x^{2} + 1; we are still looking for x = 1; it is still the same slope average formula.0716
x_{1} is still equal to 1, but now x_{2} is going to be one forward from 1, so that is 2 for our x_{2}.0723
We plug that into our average slope formula; f(x_{2}) is f(2) in this case, because that is our x_{2} at this place.0730
So, f(x_{2}) would be 2^{2} + 1; 4 + 1 gets us 5;0738
minus f(x_{1}): f(x_{1}) is still 1^{2} + 1, so that still gets us  2.0743
Our bottom is now x_{2}  x_{1}; our new x_{2} 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, nicewe are getting better; it is still not perfect, but the approximation is improving as we bring x_{2} closer to x_{1}.0766
So, as we bring our x_{2} 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 x_{2} 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 x_{2} really, really close.0787
So, let's talk about what we have been doing, in general.0793
We have our average slope formula; m_{avg} is equal to f(x_{2})  f(x_{1}), divided by (x_{2}  x_{1}).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 m_{avg} formula): we set x_{1} 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 x_{1} = 1; the first horizontal location is 1, because we want to find out about that slope.0821
Then, from there, we set x_{2} 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 x_{2} 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 x_{2} 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 x_{2} is equal to x_{1}, our starting place, plus the distance away, h.0872
So, x_{2} = x_{1} + h; with this in mind, we can now rewrite our average slope formula in terms of x_{1}.0880
We started with m_{avg} = f(x_{2})  f(x_{1}), over (x_{2}  x_{1}).0888
But now, we have this new way of writing x_{2}: x_{2} is equal to x_{1} plus the distance forward, h.0893
So, we can rewrite the formula in terms of x_{1} and the horizontal distance, h, to be able to create a new formula0898
not a new formula, so much as a restatement of the old formulabut a new way of looking at it.0906
So, we plug in x_{2} here; it now becomes x_{1} + h; so we have x_{1} + h...0911
f(x_{1}) is still the same; so f(x_{1} + h)  f(x_{1}); x_{2} now becomes x_{1} + h, minus...still x_{1}.0917
On the bottom, we have x_{1} here and x_{1} 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(x_{1} + h) and f(x_{1})...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 x_{1} part inside of the function,0947
because we have to see how h interacts with x_{1} before we can cancel anything out.0950
OK, one last thing to notice: at this point, the only thing showing up is x_{1}.0955
Only x_{1} shows up in this formula: we have x_{1} here, x_{1} here, x_{1} here, x_{1} here, x_{1} here, x_{1} here.0962
But no longer x_{2}: 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 x_{1} + h.0976
So, if we don't really care about x_{1} versus x_{2}, then we can just rename x_{1} as simply x,0978
because at heart, we are all lazy; and it is easy to write out x, compared to x_{1}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 x^{2} + 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 choosingwhatever 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 breakswhat 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 m_{avg} = [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 touchesthat 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) = x^{2} + 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) = x^{2} + 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 x^{2} + 1, so  (x^{2} + 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 x^{2} + 2xh + h^{2}.1276
The +1 still remains; and (x^{2} + 1) is x^{2}  1.1284
At this point, we see that we have positive x^{2} and negative x^{2}, 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 + h^{2} on top, all divided by h:1299
the limit as h goes to 0 of 2xh + h^{2}, 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 + h^{2}, 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 checklet'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 featherlike 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 differentiationwe 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 beforewe figured out that, in general, for x^{2} + 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 y1569
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 it1618
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 saw1677
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 caseif we end up not really using it that muchwhy 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 changingwhat 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 momentbymoment 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 changinghow 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 differentiatewhen 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 meaningto be usefulis this idea of how the function is changing here and now.1789
That is why we care about the derivativenot 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) = x^{3}, and consider the location x = 2.1850
We have some function x^{3}, 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 x^{3}, that is 0^{3}  (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 = 1so 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 1that 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 x^{3}; 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 derivativea problem about the derivative!2161
for f(x) = x^{3}, 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 x^{3}, 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 x^{3}, 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 x^{2} + 2xh + h^{2}...2239
(x + h) times x^{2} + 2xh + h^{2}...we get x^{3};2253
x times 2xh is 2x^{2}h; h times x^{2} is 1hx^{2}, or 1x^{2}h.2258
So, we get + 3x^{2}h; then, x times h^{2} is 1xh^{2}; h times 2xh is 2xh^{2};2263
so we get a total of 3xh^{2}, plus h^{3}; h^{2} times h gets us h^{3}.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 x^{3} + 3x^{2}h + 3xh^{2} + h^{3}  x^{3}, all divided by h.2293
At this point, we see that we have a positive x^{3} and a negative x^{3}; 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 h^{2} into an h^{1},2329
turns this h^{3} into an h^{2}...and we now have the limit as h goes to 0 of 3x^{2} + 3xh + h^{2}.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 h^{2} goes to 0, it will crush h^{2} to 0.2355
So, as h goes to 0, any h^{2} will crush h^{2} to 0; and that leaves us with just 3x^{2}; great.2361
So, f'(x) is equal to 3x^{2}.2368
So now, we were asked to find what is the specific derivative at f'(2).2374
f'(2) = what? Well, f'(x) = 3x^{2}, 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 x^{2} + xh.2606
Now, as h goes to 0, that will cause the xh to cancel out; but it will have no effect on the x^{2},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/x^{2} is what ends up coming out of this: 1/x^{2}.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/x^{2}; 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) = x^{n}, where n is just any constant number, the derivative of f(x) is f'(x) = n(x^{n  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) = x^{2} + 1, it came out to be simply 2x.2707
Don't be too worried about the +1; we can think of it is being x^{0}.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) = x^{3}, when we took the derivative of that in Example 2,2721
we ended up getting 3 times x^{2}: 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 x^{4}, f(x^{4}); try working through the formal definition of x^{4}.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) = x^{4} 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
x^{4} shoots up really, really quickly; we shoot up massively, very, very quickly, with x^{4}.2803
By the time we have made it to a horizontal x location of 3, we are putting out an output of 81; 3^{4} 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) = x^{4}, 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 4x^{3}.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 3^{3}; 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 slopeintercept 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 yintercept iswhat m is and what b is, at least if you are using slopeintercept 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 yvalue; 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 x^{4}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 latergoodbye!3071
0 answers
Post by Kenosha Fox on March 16, 2016
wow
0 answers
Post by James Whiddon on August 13, 2014
Excellent!
1 answer
Last reply by: Professor SelhorstJones
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 SelhorstJones
Thu Jun 6, 2013 4:25 PM
Post by HAFSA Ahmad on June 6, 2013
wow !this session is excellent.