Using the chain rule, 2017 Calculus AB free response 6a,b,c
About this video
Sal works through parts a, b and c of the 2017 AP Calculus AB free-response question 6. He uses the chain rule to find f'(pi) for f(x) = cos(2x) + e^(sin x), then the chain rule with a graph of h to find k'(pi) for k(x) = h(f(x)), and finally the product and chain rules with a table of g values to find m'(2) for m(x) = g(-2x)·h(x).
You'll be able to
- differentiate composite functions like cos(2x) and e^(sin x) with the chain rule.
- find the derivative of h(f(x)) using values from a graph.
- combine the product rule and chain rule using values from a table and a graph.
Key words
- chain rule
- The derivative of h(f(x)) is h'(f(x))·f'(x).
- slope of the tangent line
- The derivative of the function at that point.
- composite function
- A function inside another function, such as h(f(x)).
- product rule
- The derivative of a product: (uv)' = u'v + uv'.
Transcript
- [Instructor] Or to let f be the function defined by all of this stuff. Let g be a differentiable function. The table above gives the values of g and its derivative g prime at select values of x, that's that. Let h be the function whose graph consisting of five line segments as shown in the figure above.
This is h. Part a, and actually, I encourage you, just read all of this and try to solve as much of this on your own as you can before I will now work through this with you. So part a, find the slope of the line tangent to the graph of f at x equals pi. Well, the slope of the tangent line is just the derivative of f at that point.
So we just need to figure out what f prime of pi is. So the first thing we want to do is let's just figure out what f prime of x is going to be equal to. So we have to take the derivative of all of this, and both of these look like composite functions. So what we first are gonna do is take the derivative of 2x with respect to x, which is just 2, and then multiply that times the derivative of cosine of 2x with respect to 2x, so that's gonna be negative sine of 2x.
I'll put the negative out front, negative sine of 2x, and then plus, I'll put the plus out here, plus the derivative sine of x, which is cosine of x, times the derivative of e to the sine of x with respect to sine of x. So that's just going to be e to the sine of x.
So times e to the sine of x power. And so f prime at pi is going to be -2 times sine of 2 pi, sine of 2 pi plus cosine of, plus cosine of... Oh, I'm looking over here. Plus cosine of pi times e to the sine of pi. Let's see, sine of 2 pi is just going to be equal to 0, so all of that is 0.
Sine of pi is 0, so that's 0. E to the 0 power is equal to 1, so that's all 1. Cosine of pi is equal to -1. So all of this is gonna simplify to 0 plus -1 times 1, which is equal to -1. So f prime of pi is equal to -1. If you're doing this on the AP exam, I'm doing it in the margin of the problem, they're actually giving you proper space.
I encourage you to write it all out and explain your reasoning, especially if you have time to make sure that you get the full points. So that's part a, I could write it here. Remember, always show your work. F prime of pi is going to be equal to -1. Okay, part b, let k be the function defined by k of x is equal to h of f of x.
Find k prime of pi. So this looks like this whole... So far is we're gonna be doing a lot of chain rule here. So if we wanna do k prime of pi, let's just evaluate what k prime of x is going to be. K prime of x, apply the chain rule, it's gonna be the derivative of f, f prime of x, times the derivative of h with respect to f of x, which we could denote as h prime of f of x, and so then if we wanted to figure out what k prime of pi is, k prime of pi is going to be f prime of pi, which we luckily already figured out, times h prime of f of pi.
Did I have enough parentheses? Yes, I have enough parentheses. So f prime of pi, this is equal to -1, and then f of pi, we have to evaluate it up here, so if we were to say f of pi is equal to cosine of 2 pi, plus e to the sine of pi, cosine of 2 pi is equal to 1, sine of pi, we've already gone through this exercise, that's 0, e to the 0 power is, let me make sure I'm doing this right, this is e to the 0 is 1.
So we have 1 plus 1 is equal to 2, f of pi is equal to 2, and so then, so this is 2, so all of this is going to be equal to the negative of h prime of 2. So let's see, on this graph right over here, we're looking at h, this is 2, and remember, we're not looking at h of 2, we're looking at h prime of 2.
So h prime of 2 is really gonna be the slope of the line right over here. And the slope of this line is for every one it goes down or it rises down -1 or it goes -1 down and it runs or it goes three forward. And so our change in y over change in x is equal to -1 over 3 or it's equal to negative one-third.
So h prime of 2 is negative one-third, but I'm taking the negative of that, so all of this is going to be equal to the negative of negative one-third, which is equal to positive one-third, there you go. And once again, if you're doing this on the AP test, I'm trying to do it all on the margin so that we can see everything on the screen at the same time, but try to make it do it a little bit neater (laughs) if you're doing it for the graders.
Let m be the function defined by m of x is equal to all of this. Find m prime of 2. This is, again, well, this is gonna be the product rule and the chain rule, so let's get to it. Let me scroll down a little bit, get a little bit more real estate to work with.
So m prime of x is going to be equal to what we could say the derivative of this first thing, and the derivative of this first thing we have to apply the chain rule right over here. So this is going to be -2, the derivative of the inside, times the derivative of g with respect to the inside.
So just like that, yep. Times this function, so I'm also doing the product rule. So times this function, and then plus, now, I just take this first function, g of -2x, times the derivative of the second function, h prime of x. If any of this is foreign, definitely review the videos on Khan Academy on the product rule and on the chain rule.
So let's see if we can evaluate this. So m prime of 2, m prime of 2 is going to be called a -2, times g prime of -2, times two is -4, times h of 2, times h of 2, plus g of -4, -2 times 2, -4, times h prime of 2. All right, so what's this going to be? G prime of 4, let me scroll back up a little bit.
G prime when x... Oh, of -4. When x equals -4, g prime of x is -1. So g prime of -4 is, and actually, let me give myself some more space, and I had put these squiggles here to test my pen earlier, but I'm gonna need this real estate, so let me get it back. All right, so this is going to be equal to -2 times, this right over here, we just saw it was -1.
Actually, let me look at that again, I don't wanna lose track too much. So g prime of -4. When x equals -4, the derivative is -1, yes. So that's -1, and then h of 2, for h of 2, we look at this function right over here, when x is equal to 2, it looks like h...
In fact, we know because we know the slope is negative one-third, it looks like we are at y or h of 2 is equal to, h of 2 is equal to negative two-thirds. And so we could go over here, so times negative two-thirds plus g of -4, g of -4 is 5, all right? So this is equal to 5 times h prime of 2.
So the derivative at when x is equal to 2 of h, we already saw that h prime of 2, right over here, is this slope right at this point, so that is that negative one-third. So this is times negative one-third, and so all of this we get a little bit of a drum roll, I think, is going to be equal to a c.
These negatives cancel out, so we just have 2 times negative two-thirds, so that's negative four-thirds, and then minus five-thirds, minus five-thirds, is equal to negative nine-thirds, which is equal to -3, which is a nice clean number.
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