TITLE: Let $f(x)= (\sin x)^x$ Find $f'(\frac{\pi}{2})$
QUESTION [5 upvotes]: I'm not sure if I derived it right. So, Let $f(x)$ = $(\sin x)^x$. $f' = x(\cos x)$ right? then just substitute so $f' = \frac{\pi}{2}(\cos\frac{\pi}{2})$ ? Since $\cos(180°)$ is equal to $-1$. Then its -$\frac{\pi}{2}$? they said it's wrong.
after reading all the comments i arrived with $f'$ = $(sinx)^x$ $\frac{sinxln(sinx) + xcosx}{sinx}$ then substitue $\frac{pi}{2}$ to x so that's $90deg.$ $cos90$ is 0 and $sin90deg.$ is 1 so $1$ raised to $90deg.$ then $\frac {(1)(1)+0}{1}$ my answer final answer now is $\frac{pi}{2}$ which is in the multiple choice. But is it correct? This is a multiple choice by the way.

REPLY [4 votes]: Given that the problem asks for the value of $f'$ at a specific point, rather than for a general formula for $f'$, we might look for a simple solution involving geometry, rather than differentiation rules.
Note that $0\leq\sin(x)\leq1$ on $[0,\pi]$, thus $f(x)=(\sin(x))^x$ is also between $0$ and $1$ on that interval. Furthermore, $f(\pi/2) = 1$.  Thus, $f$ has a maximum at $\pi/2$ so that $f'(\pi/2)=0$.

Admittedly, I came up with this explanation after noticing that the answer was zero.  Such a simple answer, though, only begs the question further - namely, is there a simple geometric explanation.