# Derivative of Inverse Trigonometric Functions

by Shree
(Pune)

How do we derive the following function? Here we have the derivative of an inverse trigonometric function. A whole section could be devoted to this subject, but I'll try to explain this in a few words. To derive inverse functions, we just use the chain rule.

For example, in this case we have the inverse function of cotangent. This means that if we apply the inverse to cotangent, we get the initial argument.
That is: Now, if we take the derivative of both sides of this equation, we get (using the chain rule in the left side): Now, assuming that the derivative of cotangent is not zero, we can divide both sides by it to get: Now, we'd like to have a variable y in the left side instead of cot(x), so we make the change of variables: And the derivative of cotangent is: So, we have: You'll usually find this formula on most derivative tables, but it is useful to know how to deduce it. It is just the chain rule.

Now, getting back to our problem, we want to find the derivative of: And here we use the formula we just proved and the chain rule once again (plus the product rule). We can write this expression as: If we make the change of variables: We have that: Applying the chain rule: And the rest is just algebra. Do it!