Relating derivative to tangent line, 2019 Calculus AB free response 6a,b
About this video
This video works through parts (a) and (b) of the 2019 AP Calculus AB free response question 6. It uses the fact that a tangent line's slope equals the derivative at the point of tangency to find h'(2) = 2/3, then uses the product rule to find A'(x) for A(x) = 3x^3 h(x) and evaluates A'(2) = 160.
You'll be able to
- read the derivative at a point from the slope of a tangent line written in point-slope form.
- use the product rule to write an expression for the derivative of 3x^3 times h(x).
- substitute given function and derivative values to evaluate a derivative at a point.
Key words
- Twice differentiable
- A function you can take the derivative of two times, so both f' and f'' exist.
- Tangent line
- A line that touches a graph at a point and has the same slope as the graph there.
- Point-slope form
- A way of writing a line as y = y1 + m(x - x1), where m is the slope and (x1, y1) is a point on it.
- Product rule
- The derivative of f(x)g(x) is f'(x)g(x) + f(x)g'(x).
Transcript
- [Lecturer] So what we have here is the beginning of a free response question from the AP calculus exam. And this is in the non calculator section. And we're told that functions F, G and H are twice differentiable functions. Twice differentiable just means it can be differentiated twice.
With G of two is equal to H of two is equal to four. The line Y is equal to four plus two thirds times X minus two is tangent to both the graph of G at X equals two and the graph of H at X equals two. All right. So first, they ask us to find H prime of two. Pause this video, have a go at that before we do this together.
Okay, well, they tell us that at X equals two, the graph of H is tangent or the line, I should say, this line right over here is tangent to H at X equals two. So at X equals two, the derivative is going to be the same as the slope of the tangent line. So H prime of two is going to be equal to the slope of Y is equal to four plus two thirds times X minus two.
And you might immediately see you could, this is essentially an algebraically manipulated form of point slope form. And so we could say that H prime of two is equal to the slope here is two thirds. Two thirds. You could, if you're on the AP exam and you have time, it's good to really explain things out.
Maybe you could multiply, distribute the two thirds then put it in a slope intercept form or point slope form, however you wanna say it, you just have to identify that this is the slope and that this is at H prime of two is going to be the slope of the tangent line at X equals two.
B, let A be the function given by three X to the third times H of X. Write an expression for A prime of X, find A prime of two. So pause this video, have a go at it. All right, so A prime of X is going to be equal to if this is A of X, we essentially have the product of three X to the third times H of X.
So we can use the product rule. And the product rule, we take the derivative of the first thing. Maybe I'll use a couple of colors here. So I'll take the derivative of that, which is take the exponent three times the coefficient. So I get nine times x decrement the exponent three minus one is two.
So I multiply that times the second function, so times H of X. And then to that, I am going to add the first function, not taking its derivative. So three X to the third times the derivative of the second function. So times H prime of X. And so I did this first part, write an expression for A prime of X, so that's done.
And now I wanna find A prime of two. A prime of two. The derivative of A at X equals two is going to be nine times X squared. Two squared is four, times H of two, times H of two plus three times two to the third, which is eight, times H prime of two. Well, let's see, this is going to be equal to, this is 36.
H of two, we already are told is equal to four. And then this is equal to 24. And H prime of two, we just figured out what it was. It's two thirds. Two thirds. So 36 times four is 144 plus two thirds of two thirds of 24 is 16, which is going to be equal to 160.
And we're done. And you might notice, you might say, wait, they told us some things that we're not really using. We're not using the function F here, we're not using twice differentiable we're not taking a derivative twice, we're only taking it once. And that's just because this was the first part of that free response question.
I'm assuming in the later sections of this question, they are going to talk about F and maybe taking a derivative another time.
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