# Finding derivative of the inverse without the inverse

We are given a function $$f(x)=4\arcsin(\sqrt{x})+2\arcsin(\sqrt{1-x})$$ The derivative of $f$ is: $$f'(x)=\frac{1}{\sqrt{x-x^2}}$$ I would like to find the maximum value of $f^{-1}$. I think I have a solution, although I am not sure.

Solution

We know that $$\frac{d}{dx} f^{-1}(x)=\frac{1}{f'(f^{-1}(x))}$$ In the problem, we are to find when $$\frac{d}{dx} f^{-1}(x) =0$$ Thus, $$\frac{d}{dx} f^{-1}(x)= \frac{1}{\frac{1}{\sqrt{y-y^2}}}= \sqrt{y-y^2} = y(1-y)$$ It follows that $y=0$ or $y=1$, when the above expression is set to zero.

Unsure

I assume that since the inverse function gets us back to the original value, we can write this as $f(y)=x$. Is the solution I got above the maximum value or simply the point $x$ that I have to plug into $f^{-1}$ to get the maximum value?

Let $\arcsin\sqrt x=u\implies\sin u=\sqrt x\ge0\implies0\le u\le\dfrac\pi2$

$\implies\cos u=+\sqrt{1-x}$

$\implies\arcsin\sqrt{1-x}=\arcsin(\cos u)$ $=\arcsin\left[\sin\left(\dfrac\pi2-u\right)\right]=\dfrac\pi2-u$

$$\implies f(x)=4u+2\left(\dfrac\pi2-u\right)=\pi+2\arcsin\sqrt x$$

If $y=f(x),f^{-1}(y)=x$

and $y=\pi+2\arcsin\sqrt x\implies-\arcsin\sqrt x=\dfrac\pi2-\dfrac y2$

$\sin\left(-\arcsin\sqrt x\right)=\sin\left(\dfrac\pi2-\dfrac y2\right)$

$-\sin\left(\arcsin\sqrt x\right)=\cos\dfrac y2$

$\implies-\sqrt x=\cos\dfrac y2\implies x=\cos^2\dfrac y2$

$\implies f^{-1}(y)=x=\cos^2\dfrac y2$

$\implies f^{-1}(x)=\cos^2\dfrac x2=\dfrac{1+\cos x}2$

• is the maximum value of $f^{-1}$ 1? – Artem Nov 17 '14 at 16:08
• @Artem, What is $f^{-1}(x)?$ In my calculation, $$f^{-1}(x)=\cos^2\frac x2$$ – lab bhattacharjee Nov 17 '14 at 16:09
• it's the inverse of $f$. – Artem Nov 17 '14 at 16:11
• @Artem, That's obvious. Have you calculated $f^{-1}(x)?$ – lab bhattacharjee Nov 17 '14 at 16:12
• Since I only need the maximum value of $f^{-1}$, and I knew the $f$ and $f'$, I wanted to use that information to find the maximum point using the relationship in the solution. – Artem Nov 17 '14 at 16:13