find $(f^{-1})^{\prime} (\frac{ \sqrt 2 }{2})$

Solution 1:

As you said, the first thing to notice is that :

• if $0 \leq x \leq \displaystyle{\frac{\pi}{4}}$, then $\cos(x) > \sin(x)$, so $$f(x) = \int_0^x \cos(t) dt = \sin(x)$$

• if $ \displaystyle{\frac{\pi}{4} \leq x \leq \frac{\pi}{2}}$, then $\cos(x) \leq \sin(x)$, so $$f(x) = \int_0^{\frac{\pi}{4}} \cos(t) dt + \int_{\frac{\pi}{4}}^{x} \cos(t) dt = \sqrt{2} - \cos(x)$$

It is easy to see that the function $f$ is strictly increasing and differentiable from $\displaystyle{\left[ 0, \frac{\pi}{2}\right]}$ to $\displaystyle{\left[ 0, \sqrt{2}\right]}$, hence $f$ is a bijection and $f^{-1}$ is well-defined. Moreover, one has $$f \left(\frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}, \quad \quad \text{so } f^{-1} \left( \frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}.$$ Differentiating $f^{-1} \circ f = \text{Id}$, you get that $f'(x) \left(f^{-1}\right)'(f(x)) = 1$ for every $x$. In particular, for $x = \displaystyle{\frac{\pi}{4}}$, this gives $$ \left(f^{-1}\right)'\left( \displaystyle{\frac{\sqrt{2}}{2}}\right) = \frac{1}{f' \left( \displaystyle{\frac{\pi}{4}}\right)}$$

and because $f' \left( \displaystyle{\frac{\pi}{4}}\right) = \cos \left( \displaystyle{\frac{\pi}{4}}\right) = \displaystyle{\frac{\sqrt{2}}{2}}$, you get finally that $$ \boxed{ \left(f^{-1}\right)'\left( \displaystyle{\frac{\sqrt{2}}{2}}\right) = \sqrt{2}}$$