$f(x)=\sin(x)^{\sin(x)}$. Solve for extreme values of $f(x)$

Question

On finding the first derivative I get $\sin x=e^{-\sec x}$ but this doesn't solve for the critical points. Any ideas on proceeding further?

Answer 1

Hint:

Your derivative is wrong.

As a first step note that the function $y=(\sin x)^{\sin x}$ is well defined as a function $\mathbb{R}\to \mathbb{R}$ only iff $\sin x>0$, i.e.: for $2k\pi<x<(2k+1)\pi$

To find the derivative we have to write the function as:

$$ y=(\sin x)^{\sin x}=\exp(\sin x \cdot(\ln(\sin x))) $$

so, using the chain rule, the derivative is:

$$ y'= \exp(\sin x \cdot(\ln(\sin x)))\cdot\left[ \cos x \ln(\sin x)+ \frac{\cos x}{\sin x}\sin x\right]=(\sin x)^{\sin x} \cos x\left(\ln(\sin x)+1 \right) $$

from this you can find the stationary points (can you do?).

Answer 2

Let consider $h(x)=x\log(x)$, so that $f(x)=\exp(h(\sin(x)))$ for $0+2k\pi<x<\pi+2k\pi$. We have $h'(x)=\log(x)+1>0$ if and only if $x>e^{-1}$. Thus $\arcsin(e^{-1})+2k\pi$ and $\pi-\arcsin(e^{-1})+2k\pi$ are minimum, while $\frac\pi 2$ is a maximum.

Answer 3

I will work on the domain $[0,\pi]$ (defining the function at the extremes by continuity). The derivative can be written as

$$(\sin x)^{\sin x} \cos x(\ln \sin x+1).$$ We have that $f'(x)=0 \iff x\in \{\arcsin e^{-1},\pi/2,\pi-\arcsin e^{-1}\}.$ Studying the sign of the derivative:

It is negative on $(0,\arcsin e^{-1}),$ positive on $(\arcsin e^{-1},\pi/2),$ negative on $(\pi/2,\pi-\arcsin e^{-1}),$ and, positive on $(\pi-\arcsin e^{-1},\pi).$

So, there are local maximums at $x=0,x=\pi/2$ and $x=\pi.$ And, there are local minimums at $x=\arcsin e^{-1}$ and $x=\pi-\arcsin e^{-1}.$

Moreover we have $f(0)=f(\pi/2)=f(\pi)$ and $f(\arcsin e^{-1})=f(\pi-\arcsin e^{-1}).$