Let $f(x)= (\sin x)^x$ Find $f'(\frac{\pi}{2})$ 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 $f'$ = $(\sin x)^x$ $\frac{\sin x\ln(\sin x) + x\cos x}{\sin x}$ then substitute $\frac{\pi}{2}$ to x so that's $90 \deg$ $\cos90$ is $0$ and $\sin90\deg$ is $1$ so $1$ raised to $90\deg$ then $\frac {(1)(1)+0}{1}$ my final answer is now $\frac{\pi}{2}$ which is in the multiple choice. But is it correct?
 A: 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.
A: *

*Hint. Take $\log$ and then work. Please see Logarithmic Differentiation.

A: No, $f'(x)$ is not $x\cos x$. To find $f'$, first take natural logarithm of both sides of the definition of $f$:
$$
\ln(f(x)) = \ln\left((\sin x)^x\right)=x\ln(\sin x)
$$
Then take the derivative of both sides of the above equation. The left side becomes $\frac{f'(x)}{f(x)}=\frac{f'(x)}{(\sin x)^x}$, so this allows you to solve for what $f'(x)$ is. 
A: $ f'(x) = xsin(x)^{x-1}*cos(x) + (sin(x))^{x}*ln(sin(x))$
Now $f'(pi/2)$ is the limit as x goes to 0 of $f'(x)$. The first term is 0 and $sin(x)^{x}$ is 1 when x approaches 0. Therefore $f'(pi/2)$ is lim as x -> 0 of $ln(sin(x))$ which is $-\infty$.
A: Starting with $f(x)=\sin(x)^x$ note that $\ln|f(x)|=x\ln|\sin(x)|$ (the absolute values solve negativity problems) and the derivative of this is
$$\frac{f'(x)}{f(x)}=\ln|\sin(x)|+\frac {x\cos(x)}{\sin(x)}$$
Plugging in $x=\frac\pi2$, we get
$$f'\left(\frac\pi2\right)=0$$
A: use that $f(x)=(\sin(x))^x=e^{x\ln(\sin(x))}$ thus we get
$f'(x)=e^{x\ln(\sin(x))}(\ln(\sin(x))+x\frac{\cos(x)}{\sin(x)})$
