# Implicit Differentiation

So far, we have seen functions that are easy to isolate for one variable. We typically have functions in the form of f(x) = [stuff] or y = [stuff]

There are many relations and functions that can’t be rearranged in this way. Take for example the equation of a circle, . In this case, we want to be able to find the rate of change, but it isn’t easy to isolate for y. In cases like these, implicit differentiation is useful.

Example: Find the derivative of using implicit differentiation

If we start calculating the derivative of this function, here is what we get:

At this point, we need to discuss how we can compute . Since y is a function of x, we can’t just compute the derivative as if it were . Instead, we treat it like a chain rule problem. We take the derivative of the y term, then multiply the derivative of y with respect to x. Doing so gives us:

This would make our equation:

Remember, our goal is to determine what is equal to. Since this is the case, we simply need to rearrange the equation above to isolate for .

Implicit differentiation is quite easy once you get the hang of it. Every problem will run the same way. We take the derivatives, multiply by every y term, then solve for . Let’s try a few more examples.

Example: Determine y’ for

Notice that the 3xy term uses product rule. We still treat the y terms as regular functions, so derivative rules still apply. The only difference is that the y terms get the tacked on.

As you can see, implicit differentiation allows us to take derivatives of some new functions. Among them is the inverse trigonometric functions.

Example: Given , determine y’

To be able to find this derivative, we need to rearrange the equation. Recall that if , then . Knowing this, we can apply implicit differentiation.

So now, all we need to do is write cos(y) in terms of x and we are done. We stated that , so if we can rewrite cos(y) as sin(y), we can replace it with x. To do this, we use Pythagorean’s identity.

Therefore, we can substitute cos(y) to get . Since , we get:

, which is our solution.