If $f(x) = \cos x$ for all $x \in (0,\pi/2)$

Question

If $f(x) = \cos x$ for all $x\in(0,\pi/2)$, prove that $f$ is invertible and find $(f^{−1})'(1/2)$.

So i know that $f$ is invertible at $(0, \pi)$, So im wondering if $(f^{−1})'(1/2) = \cos^{-1}(1/2) =\pi/3$ ?

Answer 1

Differentiating both sides to $f\circ f^{-1}(x)=x$, we get $f'(f^{-1}(x))(f^{-1})'(x)=1$. Now plugging in $x=1/2$. Note that $f^{-1}(1/2)=\pi/3$ and $f'(u)=-\sin u$, so $f'(f^{-1}(1/2))=f'(\pi/3)=-\sin(\pi/3)=-\sqrt{3}/2$ and hence $(f^{-1})'(-1/2)=-2/\sqrt{3}$.

Answer 2

You can also use $(f^{-1})'(x) = \arccos'(x) = \frac{-1}{\sqrt{1-x^2}}$ (this can be proved) and then put $x = \tfrac{1}{2}$ to get $(f^{-1})'(\tfrac{1}{2}) =\tfrac{-2}{\sqrt{3}}$.