Exploring second derivatives (implicit equations): evaluate derivative
About this video
Sal solves a 2015 AP Calculus AB question: for y^3 - xy = 2 with dy/dx = y/(3y^2 - x), evaluate d^2y/dx^2 at (-1, 1). He uses the quotient rule with implicit differentiation, finds dy/dx = 1/4 at the point, and substitutes to get d^2y/dx^2 = 1/32.
You'll be able to
- apply the quotient rule to dy/dx = y/(3y^2 - x), using the chain rule on y terms.
- evaluate dy/dx at a point before substituting into the second derivative.
- evaluate a second derivative at a point without fully simplifying the general expression.
Key words
- quotient rule
- d/dx(f/g) = (f'g - fg')/g^2.
- implicit differentiation
- Differentiating with respect to x while treating y as a function of x.
- second derivative
- The derivative of dy/dx, written d^2y/dx^2.
Transcript
- [Instructor] So we have a question here from the 2015 AP Calculus AB test, and it says consider the curve given by the equation y to the third minus x y is equal to two. It can be shown that the first derivative of y with respect to x is equal to that. So they solve that for us.
And 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 equal negative one and y is equal to one. 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 dy, derivative of y, with respect to x is equal to y over three y squared minus x. Well if we're concerning ourselves with the second derivative, well then we wanna 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? And there's 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 derivative of y with respect to x times the denominator, three y squared minus 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?
Well the derivative of three y squared with respect to x that's going to be the derivative of three y squared with respect to y, which is just going to be six y. I'm just using the power rule there. Times the derivative of y with respect to x. All I did just now is I took the derivative of that with respect to x.
Which is derivative of that with respect to y times the derivative of y with respect to x. Comes straight out of the chain rule. Minus the derivative of this with respect to x, which is just going to be equal to one. 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 three y squared minus x, squared. Now lucky for us, they want us to evaluate this at a point, as opposed to have to do a bunch of algebraic simplification here. So we can say, when, let me do it over here.
So when, I'll it right here. When x is equal to negative one, and y is equal to one, well first of all, what's the, what's 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. The derivative of y with respect to x is going to be equal to one over three times one squared, which is just three, minus negative one.
So that's just going be plus one. It's going to be equal to 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 1/4. 1/4 times three times one squared, which is just three, minus negative one, so plus one, minus one, so I'll just leave that minus out there, times six times one, times 1/4.
Let me just write it out. Six, six times one times 1/4 minus one all of that over let's see this is going to be three times y squared, y is one, so is just gonna be three, three minus negative one, so plus one, squared. Now what is this going to be? And this is just simplifying something here.
1/4 times four that's going to simplify to one. And let's see, this is going to be 1 1/2 minus one so that's going to be 1/2. And then we're going to be have all of that over 16. And so this is going to be equal to, we get a mini drum roll here, this is going to be equal to one minus 1/2, which is equal to 1/2.
Over 16, which is the same thing as one over 32. And we are done.
Video © Khan Academy, used under CC BY-NC-SA. Watch on Khan Academy. Puzzicle is not affiliated with Khan Academy.