WEBVTT mathematics/ap-calculus-ab/hovasapian
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Hello and welcome back to www.educator.com, and welcome back to AP calculus.
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Today, we are going to start talking about ways of taking the derivative of a function, of different types of functions but doing it in a quick way.
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Let us just jump right on in.
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We have seen and we have been dealing with the definition of the derivatives.
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I will go stick with blue today.
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We have seen that the derivative of any function can be gotten from the definition of the derivative.
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That is we take the limit as h approaches 0 of f(x) + h - f(x) divided by h.
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We formed this difference quotient, we simplified it as much as possible.
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The last thing that we do is we take h to 0 to see what happens.
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We end up getting this new function.
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We need a quicker way to find derivative.
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We cannot just keep using the definition of the derivative.
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You saw from previous lessons that it can get very tedious.
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We need a quicker way, you need a quicker way to find derivatives.
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In this particular lesson, we are going to lay out the basic rules for finding derivatives of polynomial functions and exponential functions.
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For the majority of these lessons, I’m actually not going to be proving where we get these formulas
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because our main concern is having the formulas at our disposal, being able to use them.
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We want to develop technique.
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For those of you that go onto higher mathematics, you actually revisit this stuff again
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and you will spend time actually proving all of these things.
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If you want to take a look at the proofs, they are in your textbook.
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They are very straightforward, it is actually no different than anything that you have done so far.
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You are just using the definition of a derivative in its most general way.
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And then, you run through it, you prove it, they are very easy to follow.
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But again, it just tends to be a little bit tedious. We would not worry about proofs.
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The first thing we are going to talk about is the derivative of a constant, let us start there.
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The derivative of a constant.
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If you guys do not know already from previous work, the derivative with respect to the variable of a constant C is just equal to 0.
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The derivative of 5 is 0, the derivative of E is 0, the derivative of 9.7 is 0.
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Number 2, we will just call it the power rule.
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This says that if n is any real number, it can be an integer, it could be a fraction, it could be radical to, whatever, any real number.
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Then, ddx of x ⁺n is equal to n × x ⁺n-1.
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In other words, we take the n, we bring it out front, turn it into a coefficient.
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And then, we change the exponent, we drop it by 1.
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If the exponent is 5, the exponent becomes 4.
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If the exponent is 1/3, the exponent, the n-1 becomes 1/3 -3/3 becomes -2/3, that is it.
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This is the basic rule for all polynomials.
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Let us see, next rule, if C is a constant then the derivative of C × some function f(x), whatever it happens to be.
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Basically, it just says that the derivative of a constant × a function is the constant × the derivative of a function.
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We can pull out the constant and then just save it and multiply it by everything, after we take the derivative of a function, equal C × ddx of whatever f happens to be.
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I’m going to leave out the x just to make it a little bit more notationally tractable.
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Let us do the sum rule.
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If f and g are differentiable, then the derivative of f + g is just equal to,
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I’m going to leave it as it is.
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The derivative of f + the derivative of g.
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In other words, if I have some sum of a function, I can just take the derivative individually and add them up, that is it.
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The derivative of the sum is the sum of the derivatives, that is all.
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You can think of it as distributing the derivative operator over what the sum is.
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We can differentiate term by term and add.
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The last is the exponential functions.
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The exponential functions, we have the derivative of e ⁺x = e ⁺x.
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The exponential function is very special.
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When you take the derivative of it, you get the function back.
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That is very special, it is going to play a very huge role in all the mathematics that we do.
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This is going to be ddx of a ⁺x.
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In this case, we have just e, 2.718, it is this.
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Whenever there is a different constant, whatever it is, some a ⁺x, the derivative of that is,
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the a ⁺x stays but we have to multiply it by the nat-log of the base a.
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These are our basic rules for finding the derivatives of polynomial and the exponential functions.
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With that, let us just launch right into our examples because that is when everything starts to make sense.
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The first example, 9x⁵ – 4x³ + 14x² – x – 12.
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Pretty standard polynomial, 5th degree in this case.
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We have a sum, this is one polynomial but it is made up of 12345 terms.
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We are going to use the sum rule.
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We are just going to differentiate one term at a time.
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A combination of a power rule and the sum rule.
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The power rule says, take this coefficient, bring it down here, and then subtract by 1.
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Because of this 9 in front of it, it is a constant, we can pull out and save it for later.
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This is actually going to be 9 × 5x⁴ -4 ×, the 3 comes down, 4 × 3x².
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Subtract the exponent by 1, + 14, I bring the 2 down, I put here and this becomes just x¹.
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Same thing here, I bring the 1 over here and this is going to be x¹ -1 is 0, x⁰ is just 1.
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Basically, any time you are taking the derivative of 1x, 2x, 3x, 4x, it is just the number itself, the coefficient.
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The derivative of a constant is 0.
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Of course, we simplify it.
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We have 45x⁴ -12x², 14 × 2x + 28x – 1.
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There you go, that is the derivative of that polynomial, nice, straightforward, very easy.
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An application of the power rule, an application of the constant multiplied by a function, and application of the sum rule.
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When we do these, we are sort of doing them.
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We can say yes, we are applying this rule or that rule.
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After you do four of these, you are expert at it already.
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Let us look at a portion of the graph just for this particular one.
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I just wanted to take a look at it.
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What I’m going to do is we are going to go ahead and take a look at what the graph of that looks like,
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and then, the graph of the derivative on the same graph.
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Let us see, there we go.
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We have this one right here which is our f(x) that was the original function.
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And then, this right here, this is actually the derivative.
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This is f’(x).
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We have seen this before, we have dealt with it before,
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when we dealt with graphs and taking a look at the function and the derivative on the same graph.
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Notice, here the slope of this original function is positive, one is positive.
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It is positive but as it comes up here, the slope is actually decreasing.
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The line from your perspective, the slope is actually going like this, it is becoming 0.
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The slope is positive, this is the derivative.
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That is what the derivative is, it is a slope, it is positive.
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When it hits the top of the graph, it hits 0.
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Of course now the slope of this curve is negative.
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It is negative, it goes down.
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At some point, it starts to turn and starts to turn and become 0 again.
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As this point rises and it hits 0, and then from that point on, the function is rising, the slope is rising.
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The line is here, the slope is increasing and becoming positive.
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That is why you have this.
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The red is the original function, the blue is the derivative.
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It is kind of interesting, even after all these years, whenever I look at a function and its derivative on the same graph,
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it always takes me a minute to look at it and make sure I'm concentrating on the right thing.
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When you are looking at the original function, you are looking at the derivative.
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What the derivative is describing is the behavior of the slope on the original function.
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This line, that is the tangent one that is becoming that way and going down that way, and this way, and this way.
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That is what you are doing, that is what the derivative represents.
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Let us go ahead and differentiate this function, √x - 4√x.
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This radical symbol is a leftover from years and years ago, hundreds of years ago.
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I do not care for the symbol myself but again it is ubiquitous in mathematics.
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We deal with this just like turning everything into fractional exponents.
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I’m going to rewrite this f(x) as x¹/2 – x¹/4.
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I can go ahead and apply the power rule to this.
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Very simple.
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Again, this comes down, f’(x) is going to equal ½ x and ½ -1 because we subtract 1 from the exponent.
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This becomes - ½.
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This, the ¼ comes down, this becomes ¼ x.
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And then, ¼ - 1 is - ¾.
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That is it, that is your derivative.
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You are absolutely welcome to leave it like that.
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It is perfectly good mathematical symbolism.
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You can write it as follows, if you want to.
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This is really going to have to do with your teacher and what it is that they want.
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You should ask them, is it okay to leave it in this form, do you write in radical form?
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One of the things you are going to notice, if you have not noticed it, you are going to notice it now that
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we actually start do taking the derivatives of functions very quickly,
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especially when we get to product and quotient rule in the next lesson.
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There are several degrees to which you can simplify a function.
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At some point, you are just going to have to stop.
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You are going to have to ask your teacher, where it is okay to stop.
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If you want, you can write this as of 1/ 2 √x.
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This x⁻¹/2, bring it down to the denominator, and then put the radical sign back on there.
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-1/4, this is ¾, this is going to be the 4√x³.
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This is another way to write it, if you want it to.
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Example number 3, f(t) = (t/4)³.
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Let us go ahead and expand this out.
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We have f of T = t³/ 4³.
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4 × 4 is 16, 4 × 16 is 64.
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This is nothing more than 1/64 t³.
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We can go ahead and treat it, this is the constant, this is the exponent.
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Therefore, f’(t) is equal to 1/64 × 3t².
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We end up with 3/64 t² or 3t²/ 64.
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However it is that you want to write it out.
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I have always like the constant to be separate but that is just my own personal taste.
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Example number 4, x³ + 3x² + 2x + 3/ √x.
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This one again, in the next lesson we are going to be doing something called the quotient rule.
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In this particular case, we notice that there is only one thing in the dominator.
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We can actually put each of these terms over that dominator and see if it simplifies, which it does.
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I can rewrite f of x as x³ / x¹/2, because √x is x ^ ½ + 3x²/ x¹/2 + 2x/ x¹/2 + 3/ x¹/2.
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This becomes 3 - ½, this is now the same base x, I can just subtract the exponents.
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I get 3 - ½ is 2 ½, this is going to be x⁵/2 + 3x - ½ is 1 ½, 3x³/2.
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2x¹/2, 1 – ½ + 3x⁻¹/2.
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Now I have x to all these different coefficients.
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I just differentiate, f’(x) is equal to 5/2x.
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5/2 – 1 is 5/2 – 2/2 which is going to be 3/2 + 3 × 3/2x.
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3/2 – 1 is ½, ½ + 2 × ½, I bring this down.
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I subtract 1 from that ½ -1 is -½.
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And then of course, we have 3 × -1/2 x – 1/2 -1 is -3/2.
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My final answer f'(x) is equal to 5/2x³/2 + 9/2x ^½.
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The 2 and the 2 cancel.
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We are left with +x ^-½.
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This is -3/2x⁻³/2.
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There you go, that is your f’(x).
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I hope that made sense.
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F (s) is s² – 2/3√s⁴.
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We are getting pretty accustomed to dealing with this now.
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This is just s² -2.
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I will do s, this 4/3, and when I bring it up, it is going to be - 4/3.
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Is that correct, yes it is.
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We have f(s), I got to tell you that it is easy to make mistakes in differentiation.
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One of the things about calculus is it is not particularly difficult in terms of the application of the technique.
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It is just sort of keeping track of all the little things.
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In calculus, it is the details that matter, the individual little details, a - sign.
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Remembering to subtract one from the exponent.
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Remembering to add one to the exponent, coefficients, all of these things to keep track of.
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I can guarantee you, I’m going to be making my fair share of mistakes on this.
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Vigilance, that is all we can try to do is remain vigilant.
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This is going to be s⁴/3, when we bring it up -4/3.
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When we take the derivatives of this, this is going to be 2s -2 × -4/3s.
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Now, -4/3 -1 is - 7/3.
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F’ (s) = 2s, - and - becomes a +.
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2 × 4 is 8, we have 8/3s⁻⁷/3.
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There you go, this is our derivative.
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Find the equation of a tangent line to the following function at the given point.
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Graph both the function and the tangent line on the same screen.
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We have our function f(x) = x² + √x.
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We want to find the equation of a line that is the tangent line.
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In other words, that is the derivative.
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I’m sorry, no, the equation of the tangent line through this point.
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The slope of that line is going to have is the derivative of this function at that point.
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Because that is what the derivative is, the derivative is the slope of the line of the tangent line to the graph at that point.
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That was one of our interpretations of the derivative.
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The derivative is a rate of change.
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It is the rate at which y is changing, when I make a small change in x.
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It is also the slope of the tangent line to the graph at that particular x and y value, at that particular point.
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Let us go ahead and find the derivative first.
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We have f(x), let us rewrite it as x² + x¹/2.
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We have f’(x) which is going to be 2x + ½ x⁻¹/2.
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In this particular case, because I’m going to be putting values in, I’m just going to go ahead and write it in a form that is more familiar.
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+1/2 × √x.
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This is the formula for the derivative.
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When we put this x value into there, we actually get the slope of the line.
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We want to find f’(2).
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That is going to equal 2 × 2 + ½ √2.
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When I work that out, I get 4.35355.
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This is our slope, it is the slope of the tangent line.
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When I put the x value into the derivative function that I get, it gives me the slope of the tangent line at that point.
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I go ahead and find the line itself.
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The line is going to be 1 -y1 = the slope m × x - x1.
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I have x and I have y, that is the point that it passes through.
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I get y -5.414 = 4.35355 × x-2.
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I’m going to go ahead and leave it in this form.
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It is up to your teacher if they want you to actually multiply this out, simplify fractions that might show up.
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However is it that they want to see it.
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Perhaps, they want to see it in ax + by= c form, personal choice.
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That is it, nice and straightforward.
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Let us go ahead and take a look at what this actually looks like.
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This right here, this is f(x), that is the function.
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That point right there is our 2 and 5.414, that is the scales, the scales are not the same for the x and y axis.
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I sort of expanded them out.
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This is the equation of the tangent line.
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This is the tangent line, the equation that we got.
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This is the equation for the tangent line.
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This is not f’(x).
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F’(x), f’, in this particular, is f’(2).
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It is the slope of the tangent line.
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The derivative is the slope.
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The derivative is not the line itself, very important.
00:25:38.800 --> 00:25:43.900
The slope changes, if we hit peak to different point, it is going to be a different slope.
00:25:43.900 --> 00:25:48.500
If we hit to peak a point up here, it is going to be a different slope.
00:25:48.500 --> 00:25:52.300
Keep repeating it over and over again.
00:25:52.300 --> 00:26:01.100
The derivative is the slope of the tangent line not the equation of the tangent line.
00:26:01.100 --> 00:26:06.600
Find both the first and the second derivatives of e ⁺x -∛x.
00:26:06.600 --> 00:26:10.000
Very simple, we are just going to take the derivative twice.
00:26:10.000 --> 00:26:20.000
Let me rewrite f(x).
00:26:20.000 --> 00:26:28.300
F(x) = e ⁺x -, I will write this as x¹/3.
00:26:28.300 --> 00:26:34.800
We have f’(x) =, the derivative of the exponential function is the function itself.
00:26:34.800 --> 00:26:41.700
That is just e ⁺x – 1/3x⁻²/3.
00:26:41.700 --> 00:26:47.700
Bring this down, 1/3 – 1, the exponent – 1 becomes -2/3.
00:26:47.700 --> 00:26:54.300
That is f’, now I will go ahead and do f”(x).
00:26:54.300 --> 00:26:59.100
We take the derivative of the derivative, the first derivative, this becomes the second derivative.
00:26:59.100 --> 00:27:03.400
The derivative of the exponential function is e ⁺x.
00:27:03.400 --> 00:27:08.900
This is -1/3, now we take the derivative of this.
00:27:08.900 --> 00:27:21.400
We bring this down, x-2/3 – 1- 2/3- 3/3 is - 5/3.
00:27:21.400 --> 00:27:33.100
Our f’(x) and another notation for that is d ⁺2y dx², we have seen that notation before, =e ⁺x.
00:27:33.100 --> 00:27:43.600
- and - is +, 2/9x⁻⁵/3.
00:27:43.600 --> 00:27:52.200
Nice and straightforward.
00:27:52.200 --> 00:28:01.600
S = 3t³-6t² + 4t + 2 is the equation of motion of a certain particle, with s in meters and t in seconds.
00:28:01.600 --> 00:28:06.100
This is the position function.
00:28:06.100 --> 00:28:12.500
This is the position function, in other words at any time t, let us say t is 5 seconds, I put T in there.
00:28:12.500 --> 00:28:22.500
What I will end up getting is where the particle is, along the x axis tells me where it is.
00:28:22.500 --> 00:28:24.200
We want you to find the following.
00:28:24.200 --> 00:28:27.600
Find the velocity and the acceleration functions as functions of time,
00:28:27.600 --> 00:28:34.500
the acceleration after 3 seconds, the acceleration when the velocity is 0.
00:28:34.500 --> 00:28:40.800
Graph the position of the velocity and acceleration graphs on the same screen.
00:28:40.800 --> 00:28:45.000
Let us go ahead and find the velocity and acceleration functions as functions of time.
00:28:45.000 --> 00:28:47.000
Again, we have dealt with these before.
00:28:47.000 --> 00:28:51.600
Whenever you get in the position function, the velocity function is the first derivative.
00:28:51.600 --> 00:28:55.900
The acceleration function is the second derivative.
00:28:55.900 --> 00:29:07.900
Velocity = ds dt or s’, however you want to do it, that is equal to, 3 × 3 is 9.
00:29:07.900 --> 00:29:13.600
We dropped the exponent by 1, 9t².
00:29:13.600 --> 00:29:18.200
2 × 6, the 2 comes down, 2 × 6 is 12.
00:29:18.200 --> 00:29:28.200
It becomes -12t and +4 because 1 × 4 is 4 and t¹-1 is t0.
00:29:28.200 --> 00:29:29.200
It ends up going away.
00:29:29.200 --> 00:29:32.400
This is the velocity function.
00:29:32.400 --> 00:29:45.500
The acceleration that = the derivative of the velocity with respect to t, that is equal to the second derivative of the position function d² s dt² .
00:29:45.500 --> 00:29:47.800
Now I just take the derivative of the velocity.
00:29:47.800 --> 00:29:53.200
2 × 9 is 18, that is 18t-, I take the derivative of this.
00:29:53.200 --> 00:29:58.000
There we go, this is my velocity function, this is my acceleration function.
00:29:58.000 --> 00:30:03.200
At any time t, I just plug it in to get the velocity and the acceleration.
00:30:03.200 --> 00:30:10.300
At t = 2, the position is this, it is going this fast and it is accelerating with that acceleration.
00:30:10.300 --> 00:30:14.300
Nice and straightforward.
00:30:14.300 --> 00:30:20.300
Part B, they want you to find the acceleration after 3 seconds.
00:30:20.300 --> 00:30:24.300
That is nice and easy, they just want the acceleration at 3.
00:30:24.300 --> 00:30:28.800
We use this, 18 × 3 – 12.
00:30:28.800 --> 00:30:30.800
I hope to God that I did my arithmetic correctly.
00:30:30.800 --> 00:30:38.500
I’m notorious for making arithmetic mistakes but I think the answer should be 42 m/s².
00:30:38.500 --> 00:30:41.000
That is the unit of acceleration.
00:30:41.000 --> 00:30:42.800
Velocity is in meters per second.
00:30:42.800 --> 00:30:45.500
The position is in meters, velocity is in meters per second.
00:30:45.500 --> 00:30:52.600
Acceleration is in meters per second which is m/s².
00:30:52.600 --> 00:31:05.200
Part C, they say they want the acceleration when the velocity is 0.
00:31:05.200 --> 00:31:08.300
We have to set the velocity function to 0.
00:31:08.300 --> 00:31:14.600
Find the values of t at which the velocity is 0, then plug those t values back in the acceleration equation.
00:31:14.600 --> 00:31:18.600
Let us do exactly what it says.
00:31:18.600 --> 00:31:24.600
We set the V equal to 0.
00:31:24.600 --> 00:31:33.100
We have to find t, then use the acceleration function.
00:31:33.100 --> 00:31:43.000
The velocity as a function of t is equal to 9t² -12t + 4.
00:31:43.000 --> 00:31:46.700
I set that equal to 0 and I solve for t.
00:31:46.700 --> 00:31:51.600
In this particular case, I’m not going to go through the particulars of this quadratic equation.
00:31:51.600 --> 00:31:52.700
I hope that is not a problem.
00:31:52.700 --> 00:31:57.800
I'm presuming that you have your calculator at your disposal or perhaps you just want to do it by hand,
00:31:57.800 --> 00:32:01.200
by completing the square or using the quadratic formula.
00:32:01.200 --> 00:32:06.900
In this particular case, I do not think it could be factored, whatever it is that you need to do by all means.
00:32:06.900 --> 00:32:15.200
T ends up being 0.6667, one answer.
00:32:15.200 --> 00:32:25.500
When a quadratic equation has only one answer not two, that means it just touches the graph, that touches the x axis.
00:32:25.500 --> 00:32:32.300
Our acceleration that we are going to be looking for is the acceleration 0.6667.
00:32:32.300 --> 00:32:39.200
That is equal to 18 × 0.6667 -12.
00:32:39.200 --> 00:32:43.700
Our acceleration ends up being 0.
00:32:43.700 --> 00:32:53.100
Let us go ahead and take a look at the graph of all three on the same screen, on the same page.
00:32:53.100 --> 00:33:00.500
Here is our s(t), this is our position function that was the cubic equation.
00:33:00.500 --> 00:33:10.200
This one right here, the blue, this is our S‘, this is our velocity function v(t).
00:33:10.200 --> 00:33:19.100
This one right here, I think it is purple or magenta, this is S", this is the acceleration function.
00:33:19.100 --> 00:33:29.000
Notice, positive slope, velocity is positive.
00:33:29.000 --> 00:33:31.500
The velocity is positive, it is above the x axis.
00:33:31.500 --> 00:33:37.800
The particle is moving to the right.
00:33:37.800 --> 00:33:48.400
The position, the derivative actually goes to 0.
00:33:48.400 --> 00:33:55.500
The slope goes to 0 that is why it hits that and it becomes positive again, and it increases.
00:33:55.500 --> 00:34:02.600
The derivative follows the slope of the tangent line along that.
00:34:02.600 --> 00:34:05.800
The acceleration, the acceleration is the derivative of the derivative.
00:34:05.800 --> 00:34:12.400
Now, this parabola that we have, the velocity function is the function.
00:34:12.400 --> 00:34:18.100
The derivative of that is going to be a straight line.
00:34:18.100 --> 00:34:24.600
The slope is negative but it is increasing, negative but it is increasing.
00:34:24.600 --> 00:34:29.800
At this point, the slope hits 0 and then the slope become positive.
00:34:29.800 --> 00:34:34.100
That is why you have from negative to positive.
00:34:34.100 --> 00:34:43.800
The function, its first derivative, the velocity function, its second derivative, the acceleration.
00:34:43.800 --> 00:34:51.200
Nice and straightforward, I hope.
00:34:51.200 --> 00:34:55.200
Let us go ahead and try this one.
00:34:55.200 --> 00:34:57.200
A little bit longer, a little bit more involved.
00:34:57.200 --> 00:35:00.700
Again, it is not like you have a lot at your disposal.
00:35:00.700 --> 00:35:06.700
What you have at your disposal right now is this idea of the derivatives, we go with that.
00:35:06.700 --> 00:35:18.400
Find the cubic function, ax³ + bx³ + cx + d, whose graph has horizontal tangents at -2,8 and 2,0.
00:35:18.400 --> 00:35:28.100
In this particular case, we need to find a, we need to find b, we need to find c, and we need to find d, four variables that we need to find.
00:35:28.100 --> 00:35:30.300
Let us go ahead and write this out.
00:35:30.300 --> 00:35:41.700
F(x) = ax³ + bx² + cx + d.
00:35:41.700 --> 00:35:45.500
We have 4 variables.
00:35:45.500 --> 00:35:50.400
We know really only one way to find 4 variables, when we are given this little information.
00:35:50.400 --> 00:35:51.800
We need to find 4 equations.
00:35:51.800 --> 00:35:57.300
We know that if we have 4 equations and 4 unknowns, theoretically, we can find all of the unknowns.
00:35:57.300 --> 00:35:59.400
In this case, a, b, c, and d.
00:35:59.400 --> 00:36:01.100
We are going to be looking for 4 equations.
00:36:01.100 --> 00:36:04.500
Where those equations are going to come from.
00:36:04.500 --> 00:36:15.400
We know f(-2), we have it right here f(-2) is 8.
00:36:15.400 --> 00:36:23.700
We just put -2 in for x and we put 8 equal to 8.
00:36:23.700 --> 00:36:44.800
When I put -2 in for here, we get -8a + 4b, I’m putting -2 in for x, + 4b -2c + d = 8.
00:36:44.800 --> 00:36:47.400
That is my first equation.
00:36:47.400 --> 00:36:50.100
We also know f(2), f(2)=0.
00:36:50.100 --> 00:37:00.300
Again, I can put 2 in for x and get something at a, b, c, and d, that is equal to 0.
00:37:00.300 --> 00:37:12.200
We know f(2), f(2) that becomes 8a + 4b + 2c + d = 0.
00:37:12.200 --> 00:37:16.700
This is our second equation, we are doing well.
00:37:16.700 --> 00:37:17.900
They also tell us something else.
00:37:17.900 --> 00:37:22.800
They tell us that the graph actually has horizontal tangents at these points.
00:37:22.800 --> 00:37:32.800
Horizontal tangent means that the graph itself, no matter what the graph looks like, still the slope is 0.
00:37:32.800 --> 00:37:36.500
Therefore, the slope is the derivative.
00:37:36.500 --> 00:37:46.200
I take the derivative of this thing to get a new equation and I set that equal to 0 at these points.
00:37:46.200 --> 00:37:52.200
I plug in -2 and 2 into the derivative of this to get two other equations.
00:37:52.200 --> 00:37:55.700
Let us write all that out.
00:37:55.700 --> 00:38:17.600
Horizontal tangents at x = -2 and x =2, that means a slope of 0.
00:38:17.600 --> 00:38:31.300
This implies that f’ of x at 2 and -2 = 0.
00:38:31.300 --> 00:38:34.200
A slope of 0 means the derivative is 0.
00:38:34.200 --> 00:38:36.500
Let us take f’(x).
00:38:36.500 --> 00:38:51.300
F’(x) is equal to 3ax² + 2bx + c.
00:38:51.300 --> 00:38:54.200
The derivative of d is 0.
00:38:54.200 --> 00:38:59.600
Now let us find, let us form the equation f’(-2).
00:38:59.600 --> 00:39:04.000
F’(-2), -2 we are putting into the x value.
00:39:04.000 --> 00:39:08.300
-2² is 4, 4 × 3 is 12.
00:39:08.300 --> 00:39:10.500
We have 12a.
00:39:10.500 --> 00:39:15.100
-2 and 2 this is -4b + c.
00:39:15.100 --> 00:39:19.700
We know that the slope is actually equal to 0.
00:39:19.700 --> 00:39:22.000
We have our third equation.
00:39:22.000 --> 00:39:24.500
Now let us do f’ at 2.
00:39:24.500 --> 00:39:31.400
F’ at 2 is going to be 12a + 4b + c.
00:39:31.400 --> 00:39:34.500
That slope is also equal to 0.
00:39:34.500 --> 00:39:36.500
There we have it.
00:39:36.500 --> 00:39:43.900
We have our first equation, we have our second equation, we have our third equation, and we have our fourth equation.
00:39:43.900 --> 00:39:50.800
4 variables, 4 equations, 4 equations, 4 unknowns.
00:39:50.800 --> 00:39:53.600
Let us go ahead and work out the rest here.
00:39:53.600 --> 00:39:56.500
We have 4 equations and 4 unknowns.
00:39:56.500 --> 00:40:01.600
Our unknowns are a, b, c, and d.
00:40:01.600 --> 00:40:05.000
Let us rewrite the equations again.
00:40:05.000 --> 00:40:20.400
We have -8a + 4b -2c + d, that = 8.
00:40:20.400 --> 00:40:22.700
That is fine, I guess I can do it over here.
00:40:22.700 --> 00:40:34.700
We have the equation 8a + 4b + 2c + d = 0.
00:40:34.700 --> 00:40:41.000
We have 12a -4b + c = 0.
00:40:41.000 --> 00:40:47.800
We have 12a + 4b + c = 0.
00:40:47.800 --> 00:40:52.200
Again, I’m not going to go through the process of actually solving these equations simultaneously.
00:40:52.200 --> 00:40:56.400
It just has to do with, you can enter it into your software.
00:40:56.400 --> 00:40:58.200
You can do it on your calculator.
00:40:58.200 --> 00:41:06.500
You can do it at any number of online free equation solvers, that is how I did it.
00:41:06.500 --> 00:41:09.500
Or you can do it by hand, I also did it by hand just to double check.
00:41:09.500 --> 00:41:15.300
By fiddling around with some equations, solving for one variable here, solving for the same variable here.
00:41:15.300 --> 00:41:18.800
Putting into these equations, whatever technique that you are going to use.
00:41:18.800 --> 00:41:21.900
Again, these are techniques from previous mathematics.
00:41:21.900 --> 00:41:26.800
We are just using them in support of what it is that we are doing now, with just calculus.
00:41:26.800 --> 00:41:36.000
However you solve this, when you do, you end up with a = 0.1667.
00:41:36.000 --> 00:41:38.000
You end up with b = 0.
00:41:38.000 --> 00:41:40.800
You end up with c = -2.
00:41:40.800 --> 00:41:47.400
You end up with d = 5.333.
00:41:47.400 --> 00:42:12.800
Our f(x) equation, the equation that we wanted is 0.1667 x³ - 2x is +0x², I will just leave that out, +5.333.
00:42:12.800 --> 00:42:14.900
There you go, nice and straight forward.
00:42:14.900 --> 00:42:23.100
Again, we have the functions, we have the points, the x and y values,
00:42:23.100 --> 00:42:26.300
and they say horizontal tangent so we know that the derivative is 0.
00:42:26.300 --> 00:42:33.700
We can create 4 equations in those 4 unknowns.
00:42:33.700 --> 00:42:41.700
Let us see, find the points on the graph of the function f(x) = 1 + 3 e ⁺x-2x
00:42:41.700 --> 00:42:50.000
with tangent line at such a point is parallel to the line, the 2x – y = 4.
00:42:50.000 --> 00:42:59.100
We have f(x) that is equal to 1 + 3 e ⁺x – 2x.
00:42:59.100 --> 00:43:05.700
Let us go ahead and take the derivative of this, point at the tangent line
00:43:05.700 --> 00:43:11.000
because the slope of the tangent line is going to be the derivative at that particular point.
00:43:11.000 --> 00:43:19.900
F'(x) = 0, it is going to be 3 e ⁺x-2.
00:43:19.900 --> 00:43:28.700
They say that the tangent line, the tangent line at such a point is parallel to the line 2x – y = 4.
00:43:28.700 --> 00:43:35.500
Over here, I have the line 2x - y = 4.
00:43:35.500 --> 00:43:41.800
This is y = 2x-4.
00:43:41.800 --> 00:43:46.700
If the tangent line is going to be parallel to this line, it is going to have the same slope.
00:43:46.700 --> 00:43:52.800
It is going to have a slope of 2.
00:43:52.800 --> 00:43:57.100
The f'(x) is an equation for the slope.
00:43:57.100 --> 00:44:02.900
Therefore, I’m going to set this equal to 2.
00:44:02.900 --> 00:44:15.100
F'(x), we want f'(x), the slope of the tangent line to equal 2 because it is parallel.
00:44:15.100 --> 00:44:21.400
Let us form that equation.
00:44:21.400 --> 00:44:23.200
Let me do it over here.
00:44:23.200 --> 00:44:31.700
We have 3 e ⁺x-2 = 2.
00:44:31.700 --> 00:44:38.600
We get 3 e ⁺x = 4/3.
00:44:38.600 --> 00:44:41.100
We take the nat-log of both sides.
00:44:41.100 --> 00:44:50.800
We get x = nat-log of 4/3.
00:44:50.800 --> 00:45:00.500
When we do this, the nat-log of 4/3 is 0.2877, that is the x value.
00:45:00.500 --> 00:45:03.400
We take this x value.
00:45:03.400 --> 00:45:09.400
They want the point on the graph of the function where the tangent line at such a point is parallel to the line.
00:45:09.400 --> 00:45:11.700
We found the x value of that point.
00:45:11.700 --> 00:45:17.100
If we want the y value, we have to put it back into the original function.
00:45:17.100 --> 00:45:32.100
We get f(0.2877) that is equal to 4.247.
00:45:32.100 --> 00:45:49.000
Our point is 0.2877, 4.247, there we go.
00:45:49.000 --> 00:45:54.900
Once again, find the point on the graph of the function f(x) where the tangent line at such a point is parallel to the line.
00:45:54.900 --> 00:45:57.800
I took the derivative of this function.
00:45:57.800 --> 00:46:06.900
The derivative of that function at a given point is going to be a slope of the tangent line through that point.
00:46:06.900 --> 00:46:14.900
They say that the line, the tangent line is parallel to this.
00:46:14.900 --> 00:46:16.900
The slope of this line is 2.
00:46:16.900 --> 00:46:19.800
I set the derivative equal to 2 to find the x value.
00:46:19.800 --> 00:46:25.800
I found this x value, I plugged it back in the original equation to find the y value.
00:46:25.800 --> 00:46:31.200
It passes through this point and has a slope of 2.
00:46:31.200 --> 00:46:33.500
Thank you so much for joining us here at www.educator.com.
00:46:33.500 --> 00:47:35.000
We will see you next time, bye.