# Derivative of inverse trig function

I'm a bit confused about how the inverse trig functions are differentiated. From a website:

We have the following relationship between the inverse sine function and the sine function.

$$\sin \left( {{{\sin }^{ - 1}}x} \right) = x\hspace{0.5in}{\sin ^{ - 1}}\left( {\sin x} \right) = x$$

In other words they are inverses of each other. This means that we can use the fact above to find the derivative of inverse sine. Let’s start with,

$$f\left( x \right) = \sin x\hspace{0.5in}g\left( x \right) = {\sin ^{ - 1}}x$$

Then,

$$g'\left( x \right) = \frac{1}{{f'\left( {g\left( x \right)} \right)}} = \frac{1}{{\cos \left( {{{\sin }^{ - 1}}x} \right)}}$$

This is not a very useful formula. Let’s see if we can get a better formula. [..]

Why is the author saying this is not a very useful formula and we need to continue further? It looks like the derivative of $$g'(x)$$ is expressed in terms of $$x$$ which is what we want and we applied the inverse function rule correctly.

Is there a reason for continuing until we arrive at:

$$\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \frac{1}{{\sqrt {1 - {x^2}} }}$$

## 1 Answer

They are both correct and they are equal to each other, but $${\sqrt {1 - {x^2}} }$$ is much easier to compute and read than $${\cos \left( {{{\sin }^{ - 1}}x} \right)}$$.