# Finding Second Derivative of Implicit Function

by Laura

This is an example of a more elaborate implicit differentiation problem. This is an interesting problem, since we need to apply the product rule in a way that you may not be used to. We're asked to find y'', that is, the second derivative of y with respect to x, given that:

We apply the derivative operator to both sides and the chain rule:

Because the derivative of 1 is zero. Now, the first term is a product, so, we apply the product rule. For the second term we apply the chain rule:

Now, since x' = 1 (we're taking the derivative with respect to x), we have that:

Now, we need to apply the derivative operator again:

We now apply the product rule in the first and third term:

Now, we just need to solve for y'':

Now, it would be nice if we replace that y' in there for an expression in x and y. We can do that solving for y' in the equation we got when we differentiated for the first time:

Now, replacing this in our expression for y'':

The rest is algebra (and a bit boring), so I leave it to you. :)