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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.

All right, so let's just go to the beginning where they tell us that d y d x is equal to y over three y squared minus x. So let me write that down.

So d y d x is equal to y over three y squared minus x, three y squared minus x.

Now, what I want to 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 going to have to apply the quotient rule, which 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 if I multiply both sides of this times 3y squared minus x, then I get (3y squared minus x) times d y d x 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.

So let's apply our derivative operator to both sides. Let's give ourselves some space. So I'm going to apply my derivative operator to both sides, take the derivative of the left-hand side with respect to x and the derivative of the right-hand side with respect to x.

Here, I'll apply the product rule first. I'll take the derivative of this expression and then multiply that times d y d x, and then I'll take the derivative of this expression and multiply it times this.

The derivative of (3y squared minus x) with respect to x, well that is going to be—so if I take the derivative of 3y squared with respect to y, it's going to be 6y, and then I have to take the derivative of y with respect to x, so times d y d x, as we say. If this is completely unfamiliar to you, I encourage you to watch several videos on Khan Academy on implicit differentiation, which is really just an extension of the chain rule.

I took the derivative of 3y 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 1. Fair enough.

So I took the derivative of this; I'm going to multiply it times that, so times d y d x. 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 did is apply the product rule: (3y squared minus x) once again; I'll say, the third time: the derivative of this times this plus the derivative of this times that.

All right, that's going to be equal to, on the right-hand side, the derivative of y with respect to x is just d y d x. Now, if I want to, I could solve for d d—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 want to 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 x's here? And what's d y d x? Well, d y d x 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, I think that's where we evaluated in the beginning—yep, at the point (-1, 1). At the point (-1, 1), so that's d y d x is one-fourth.

So this is one-fourth, this is one-fourth, and this is one-fourth. Now we can solve for the second derivative—just going to make sure I don't make any careless mistakes.

So six times one times one-fourth—four is going to be 1.5, or 6. 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, plus two times the second derivative of y with respect to x is equal to one-fourth.

And so let's see; I have 1/8 plus 2 times the second derivative is equal to 1/4. And let's see, I could subtract 1/8 from both sides.

So subtract 1/8—subtract 1/8. These cancel, I get 2 times the second derivative of y with respect to x is equal to 1/4, the same thing as 2/8. So 2/8 minus 1/8 is 1/8.

Then I can divide both sides by 2, and we get a little bit of a drum roll here. You divide both sides by two, or you could say multiply both sides by one-half, and 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 sixteen. And we're done.