# Differentiating inverse functions

Functions such as arcsin(x), arccos(x), arctan(x), etc. are considered to be inverse functions of sin(x), cos(x) and tan(x) respectively. They have their own properties and as such, they have derivatives which are noticeably different from their ordinary counterparts. In my previous article: https://medium.com/@oscarnieves100/implicit-differentiation-in-a-nutshell-46031531f34b I discussed how to use implicit differentiation to find derivatives of functions which cannot be expressed in the standard form y = f(x). In this article, I will explain how to use that same technique to differentiate inverse functions.

First, consider y = arccos(x). We know nothing about the derivative of arccos(x), but we know how to differentiate cos(x), so let us rewrite this as

Next, we use implicit differentiation and the chain rule on the left-hand side to obtain

which upon re-arranging yields:

which is the derivative of arccos(x). This seems fine, but we can simplify things further. Consider the trigonometric identity:

Evaluating sin(arccos(x)) using this substitution then yields:

because cos(arccos(x)) = x. Hence, we derive the following result:

This can be extended further. Consider a more general inverse function

which rearranged looks like

Then via implicit differentiation we obtain

and then

Using the same identity for sin(arccos(x)) we used in the previous example, we can then write this as

which is valid for an arbitrary function f(x). Using this technique, it is also easy to show that the following two results are true:

Similarly, we can use this technique for other inverse functions, like the inverse csc function:

or simply

The left-hand side is actually equivalent to

which leads to the following result:

and so

This can be simplified further via the trigonometric identity:

which can be rewritten as

and therefore