# What is the derivative of f(x)=ln(sin^-1(x)) ?

##### 1 Answer
Jul 24, 2014

A side comment to start with: the notation ${\sin}^{-} 1$ for the inverse sine function (more explicitly, the inverse function of the restriction of sine to $\left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$) is widespread but misleading. Indeed, the standard convention for exponents when using trig functions (e.g., ${\sin}^{2} x : = {\left(\sin x\right)}^{2}$ suggests that ${\sin}^{- 1} x$ is ${\left(\sin x\right)}^{- 1} = \frac{1}{\sin x}$. Of course, it is not, but the notation is very misleading. The alternative (and commonly used) notation $\arcsin x$ is much better.

Now for the derivative. This is a composite, so we will use the Chain Rule. We will need $\left(\ln x\right) ' = \frac{1}{x}$ (see calculus of logarithms ) and $\left(\arcsin x\right) ' = \frac{1}{\sqrt{1 - {x}^{2}}}$ (see calculus of inverse trig functions ).
Using the Chain Rule:
$\left(\ln \left(\arcsin x\right)\right) ' = \frac{1}{\arcsin} x \setminus \times \left(\arcsin x\right) ' = \frac{1}{\arcsin x \sqrt{1 - {x}^{2}}}$.