### Derivatives of inverse functions (video) | Khan Academy

The graphs of a function and its inverse are mirror images across the line y = x. . zero, as we would in taking a derivative, so the relationship between derivatives is: (You will likely encounter problems like this on the AP calculus exams). One very important application of implicit differentiation is to finding deriva tives of inverse functions. How are the graphs of a function and its inverse related? We start by graphing f(x) SC Single Variable Calculus . Fall In mathematics, the inverse of a function y = f (x) {\displaystyle y=f(x)} y=f(x) is a function that, and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at x {\displaystyle x} x . Integrating this relationship gives . Calculus · Inverse functions · Chain rule · Inverse function theorem · Implicit.

In most cases either is acceptable. For the two functions that we started off this section with we could write either of the following two sets of notation.

The process for finding the inverse of a function is a fairly simple one although there are a couple of steps that can on occasion be somewhat messy. This is done to make the rest of the process easier.

## 3.7: Derivatives of Inverse Functions

This is the step where mistakes are most often made so be careful with this step. This work can sometimes be messy making it easy to make mistakes so again be careful.

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- The Derivative of an Inverse Function

Most of the steps are not all that bad but as mentioned in the process there are a couple of steps that we really need to be careful with since it is easy to make mistakes in those steps. For all the functions that we are going to be looking at in this course if one is true then the other will also be true. However, there are functions they are beyond the scope of this course however for which it is possible for only one of these to be true.

This is brought up because in all the problems here we will be just checking one of them. We just need to always remember that technically we should check both. However, it would be nice to actually start with this since we know what we should get. This will work as a nice verification of the process.

Here are the first few steps.

The next example can be a little messy so be careful with the work here. With this kind of problem it is very easy to make a mistake here. If a point x, y exists on f xthen the point y, x is its mirror image on f-1 x.

## Inverse functions and differentiation

For example, 2, 4 is on the graph of f x below, and 4, 2 is its mirror image on f-1 x. This mirror-image property will help us a lot as we take derivatives of inverse functions.

The derivative of an inverse function Now let's take a generic function, f xand its inverse f-1 x. We begin with the statement Now we differentiate both sides with respect to x, remembering that we need to apply the chain rule to the composition of functions on the left: If we divide to isolate the derivative of the inverse on the left, we finally get: Derivative of an inverse function The derivative of an inverse function, f-1 x can be found without directly taking the derivative, if we know the function, f xand its derivative.

Example 1 Derivatives of the inverse trig. We can build a triangle that reflects that fact: To get the bottom side, just use the Pythagorean theorem. Now we can replace cos y with an algebraic expression containing x: Now taking the derivative of both sides gives: Derivative of cos-1 x This derivative is calculated in much the same way.

### Calculus I - Inverse Functions

We'll skip the details for this one; you should try it on your own. You could use the same method to find derivatives of the inverse cosecant, secant and cotangent functions, too.

Example 2 Show that the derivative of sin-1 xas obtained above, is equivalent to the derivative found using Solution: We begin with our function and its inverse: We need the derivative of the function Then it's just a matter of plugging the inverse in to cos x: Now it's a little difficult to convert this into the form we found in example 1, but if we plot the two, the result is in the graph on the right. I've shifted the purple curve upward by 0.