Second derivatives (implicit equations): evaluate derivative | AP Calculus AB | Khan Academy

So we have a question here from the 2015 AP Calculus AB test, and it says, "Consider the curve given by the equation ( y^3 - xy = 2 )." It can be shown that the first derivative of ( y ) with respect to ( x ) is equal to that. So they solved that for us.

Then part C of it, I skipped parts A and B for the sake of this video: Evaluate the second derivative of ( y ) with respect to ( x ) at the point on the curve where ( x = -1 ) and ( y = 1 ).

So pause this video and see if you can do that.

All right, now let's do it together. And so let me just first write down the first derivative. So ( \frac{dy}{dx} = \frac{y}{3y^2 - x} ).

Well, if we're concerning ourselves with the second derivative, then we want to take the derivative with respect to ( x ) of both sides of this. So let's just do that. Do the derivative operator on both sides right over here.

Now, on the left-hand side, we of course are going to get the second derivative of ( y ) with respect to ( x ). But what do we get on the right-hand side? There are multiple ways to approach this, but for something like this, the quotient rule probably is the best way to tackle it.

I sometimes complain about the quotient rule, saying, "Hey, it’s just a variation of the product rule," but it's actually quite useful in something like this. We just have to remind ourselves that this is going to be equal to the derivative of the numerator with respect to ( x ), and so that's just going to be ( \frac{dy}{dx} ) times the denominator ( (3y^2 - x) ) minus the numerator ( (y) ) times the derivative of the denominator with respect to ( x ).

Well, what's the derivative of this denominator with respect to ( x )? The derivative of ( 3y^2 ) with respect to ( x ) is going to be the derivative of ( 3y^2 ) with respect to ( y ), which is just going to be ( 6y ) (I’m just using the power rule there) times the derivative of ( y ) with respect to ( x ). All I did just now is take the derivative of that with respect to ( x ), which is the derivative of that with respect to ( y \times \frac{dy}{dx} ) come straight out of the chain rule minus the derivative of this with respect to ( x ), which is just going to be equal to ( 1 ).

All of that over—remember we’re in the middle of the quotient rule right over here—all of that over the denominator squared. All of that over ( (3y^2 - x)^2 ).

Now lucky for us, they want us to evaluate this at a point, as opposed to having to do a bunch of algebraic simplification here. So we can say when—let me do it over here—so when ( x = -1 ) and ( y = 1 ).

Well, first of all, what's ( \frac{dy}{dx} ) going to be? The derivative of ( y ) with respect to ( x )—let me scroll down a little bit so we have a little bit more space—is going to be equal to ( \frac{1}{3 \cdot 1^2} ) which is just ( 3 - (-1) ).

So that’s just going to be plus 1; it's going to be equal to ( \frac{1}{4} ).

And so this whole expression over here, so I can write the second derivative of ( y ) with respect to ( x ) is going to be equal to—well we know that—that's going to be equal to (\frac{1}{4} \cdot 3 \cdot 1^2) which is just ( 3 - (-1) ) so plus 1 minus 1.

So I’ll just leave that minus out there, times ( 6 \cdot 1 \cdot \frac{1}{4} ). Let me just write it out: ( 6 \cdot 1 \cdot \frac{1}{4} ) minus 1.

All of that over—let's see—this is going to be ( 3 \cdot y^2 ) where ( y = 1 ). So this is going to be ( 3 \cdot (3 - (-1)) ) so plus 1 squared.

Now, what is this going to be? This is just simplifying something here: ( \frac{1}{4} \cdot 4 ), that's going to simplify to 1. And let’s see, this is going to be one and a half minus 1, so that's going to be ( \frac{1}{2} ) and then we're going to have all of that over 16.

And so this is going to be equal to—well, get a mini drum roll here—this is going to be equal to ( 1 - \frac{1}{2} ) which is equal to ( \frac{1}{2} ) over 16, which is the same thing as ( \frac{1}{32} ).

And we are done.