Question 2.8.5: Finding the Derivative of an Inverse Trigonometric Function ...

From the chain rule, we have

(a) \frac{d}{d x} \cos ^{-1}\left(3 x^2\right)=\frac{-1}{\sqrt{1-\left(3 x^2\right)^2}} \frac{d}{d x}\left(3 x^2\right)

=\frac{-6 x}{\sqrt{1-9 x^4}}.

(b) \frac{d}{d x}\left(\sec ^{-1} x\right)^2=2\left(\sec ^{-1} x\right) \frac{d}{d x}\left(\sec ^{-1} x\right)

=2\left(\sec ^{-1} x\right) \frac{1}{|x| \sqrt{x^2-1}}

and (c) \frac{d}{d x}\left[\tan ^{-1}\left(x^3\right)\right]=\frac{1}{1+\left(x^3\right)^2} \frac{d}{d x}\left(x^3\right)

=\frac{3 x^2}{1+x^6}.