# What is the derivative of f(x)=cos^-1(x^3) ?

A side comment to start with: the notation ${\cos}^{-} 1$ for the inverse cosine function (more explicitly, the inverse function of the restriction of cosine to $\left[0 , \pi\right]$) is widespread but misleading. Indeed, the standard convention for exponents when using trig functions (e.g., ${\cos}^{2} x : = {\left(\cos x\right)}^{2}$ suggests that ${\cos}^{- 1} x$ is ${\left(\cos x\right)}^{- 1} = \frac{1}{\cos x}$. Of course, it is not, but the notation is very misleading. The alternative (and commonly used) notation $\arccos x$ is much better.
Now for the derivative. This is a composite, so we will use the Chain Rule. We will need $\left({x}^{3}\right) ' = 3 {x}^{2}$ and $\left(\arccos x\right) ' = - \frac{1}{\sqrt{1 - {x}^{2}}}$ (see calculus of inverse trig functions ).
$\left(\arccos \left({x}^{3}\right)\right) ' = - \frac{1}{\sqrt{1 - {\left({x}^{3}\right)}^{2}}} \setminus \times \left({x}^{3}\right) ' = - \frac{3 {x}^{2}}{\sqrt{1 - {x}^{6}}}$.