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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition) > Unit 8

Lesson 1: AP Calculus AB questions- Sal interviews the AP Calculus Lead at College Board
- 2015 AP Calculus AB/BC 1ab
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- 2015 AP Calculus AB 6a
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# 2015 AP Calculus AB 6c

Finding second derivative through implicit differentiation.

## Want to join the conversation?

- I got 1/32. Did anyone else get that?(12 votes)
- At5:36isn't the second derivative supposed to be multiplied by 4? 3- (-1) = 4 not two...I could be mistaken in my method however.(5 votes)
- Is there a video that goes through the steps of part b?(4 votes)
- SAL YOU GOT IT WRONG AT5:13. IT'S NOT 2 times second derivative, it's 4 times second derivative!(4 votes)
- 0:00Where can I find the explanation of 6b?(1 vote)

## Video transcript

- [Voiceover] Part C, evaluate
the second derivative of y with respect to x squared at the point on the curve
where x equals negative one and y is equal to one. Alright, so let's just go to the beginning where they tell us that dy dx is equal to y over three y squared minus x. So let me write that down. So dy dy dx is equal to y over three y squared minus x. Three y squared minus x. And now what I wanna do is essentially, take the derivative of both sides again. Now, there's a bunch of ways
that we could try to tackle it, but the way it's written right now, if I take the derivative
of the right-hand side, I'm gonna have to apply the quotient rule, which is really just comes
from the product rule, but it gets pretty hairy. So, let's see if I can
simplify my task a little bit. I'm going to have to do
implicit differentiation one way or the other, so what if I multiply both sides of this times three y squared minus x? Then I get three y squared minus x times dy dx is equal to, is equal to y. And so now, taking the
derivative of both sides of this is going to be a little
bit more straightforward. In fact, if I want, I could, well actually, let me
just do it like this. And so, let's, let's apply our derivative
operator to both sides. Let's give ourselves some space. So I'm gonna apply my derivative
operator to both sides. We take the derivative
of the left-hand side with respect to x, and the derivative of the right-hand side with respect to x. And so, here I'll apply the product rule. First, I'll take the
derivative of this expression and then multiply that times dy dx, and then I'll take the
derivative of this expression and multiply it times this. So, the derivative of
three y squared minus x with respect to x, well, that is going to be, so if I take the derivative
of three y squared with respect to y, it's going to be six y, and then I have to take
the derivative of y with respect to x, so times dy dx. As we say, and this is completely unfamiliar to you and I encourage you to
watch the several videos on Khan Academy on
implicit differentiation, which is really just an
extension of the chain rule. I took the derivative of three y squared with respect to y, and took the derivative of y with respect to x. Now, if I take the
derivative of negative x with respect to x, that's going to be negative one. Fair enough. And so, I took the derivative of this, I'm gonna multiply it times that. So times dy dx and then to that, I'm going to add the derivative of this with respect to x. Well, that's just going to
be the second derivative of y with respect to x times this. And all I do is apply the product rule. Three y squared minus x. Once again, I'll say it a third time, derivative of this times this plus the derivative of this times that. Alright. That's going to be equal to, on the right-hand side, derivative of y with respect to x is just dy dx. Now, if I want to, I could solve for the
second derivative of y with respect to x, but what could be even better than that is if I substitute
everything else with numbers because then it's just going to be a nice, easy numerical equation to solve. So, we know that, what we wanna figure out, our second derivative, when y is equal to one. So y is equal to one. So that's going to be equal to one and this is going to be equal to one. X is equal to negative one. So, let me underline that. X is equal to negative one. So x is negative one. Are there any other xs here? And what's dy dx? Well, dy dx, dy dx is going to be equal to one over three times one, which is three, minus negative one. So, this is equal to one fourth. In fact, that's I think where
we evaluated in the beginning. Yep, at the point negative one comma one. At the point negative one comma one. So that's dy dx is one fourth. So this is one fourth, this is one fourth, and this is one fourth. And now, we can solve for the second can solve for the second derivative. Just gonna make sure I don't
make any careless mistakes. So if six times one times one fourth is going to be one point five, or six, well, six times one times
one fourth is six fourths, minus one is going to be, so six fourths minus four
fourths is two fourths, so this is going to be one half times one fourth, times one fourth, that's this over here, plus the second derivative of y with respect to x, and here I have one minus negative one, so it's one plus one, so times two. So maybe I'll write it this way. Plus two times the second derivative of y with respect to x is equal to one fourth. And so let's see, I have one eighth plus two times the second derivative is equal to one fourth. And let's see, I could subtract one
eighth from both sides. So let's subtract one eighth. Subtract one eighth. These cancel. I get two times the second derivative of y with respect to x is equal to, one fourth is the same
thing as two eighths, so two eighths minus one
eighth is one eighth. And then I can divide both sides by two, and we get a little bit
of a drum roll here, we divide both sides by two, or you could say multiple
both sides times one half, and you are going to get, you are going to get the second derivative of y with respect to x when x is negative one and y is equal to one, is one over 16. And we're done.