Derivative with trig functions and ln trickery I am suppose to differentiate 
$y=(\sin x)^{\ln x}$
I have absolutely no idea, this was asked on a test and I just do not know how to do this I have forgotten the tricks I was suppose to memorize for the test.
 A: Hint 
$$\sin x=e^{\ln \left( \sin x\right) }\Rightarrow \left( \sin x\right) ^{\ln x}=\left( e^{\ln \left( \sin x\right) }\right) ^{\ln x}=e^{\left( \ln x\right) \;\cdot\;\ln \left( \sin x\right) }\tag{1}$$
and evaluate the derivative of $e^{\left( \ln
x\right) \ln \left( \sin x\right) }.$
Comments (trying to reply to OP's comments). 


*

*We can start by writing the given function as $$y=(\sin x)^{\ln x}=e^{\left( \ln x\right) \;\cdot\;\ln \left( \sin x\right)},\tag{2}$$ which is a particular case of the algebraic identity $$\left[ u\left( x\right) \right] ^{v\left( x\right) }=e^{v(x)\;\cdot\;\ln(u(x))}.\tag{3}$$ Remarks.  We've used the following properties. By the definition of the natural logarithm, we have (see Powers via logarithms) $$\ln u=v\Leftrightarrow u=e^v=e^{\ln u},\tag{4}$$  and the rule $(a^b)^c=a^{b\;\cdot\; c}\tag{5}$  

*Finally we evaluate the derivative of $e^{g(x)}$, where $$g(x)=\left( \ln x\right)\;\cdot\; \ln \left( \sin x\right)\tag{6}.$$ By the chain rule we have $$y'=(e^{g(x)})'=e^{g(x)}g^{\prime }(x),\tag{7}$$
and $g'(x)$ is to be computed by the product rule.


Evaluation of $(7)$
$$\begin{eqnarray*}
g^{\prime }(x) &=&\left( \left( \ln x\right) \ln \left( \sin x\right)
\right) ^{\prime } \\
&=&\left( \ln x\right) ^{\prime }\ln \left( \sin x\right) +\left( \ln
x\right) \left( \ln \left( \sin x\right) \right) ^{\prime } \\
&=&\frac{1}{x}\ln \left( \sin x\right) +\left( \ln x\right) \frac{\cos x}{
\sin x} \\
&=&\frac{\ln \left( \sin x\right) \sin x+\left( \left( \ln x\right) \cos
x\right) x}{x\sin x}.\tag{8}
\end{eqnarray*}$$
Hence, since $e^{g(x)}=\left( \sin x\right) ^{\ln x}$, we obtain 
$$\begin{eqnarray*}
y^{\prime } &=&e^{g(x)}g^{\prime }(x)=y\;\cdot\; g^{\prime }(x) \\
&=&\left( \sin x\right) ^{\ln x}\frac{\ln \left( \sin x\right) \sin x+\left(
\left( \ln x\right) \cos x\right) x}{x\sin x}.\tag{9}
\end{eqnarray*}$$
A: You can either use the definition that says that
$$a^b = e^{b\ln(a)}$$
and use the chain rule with
$$y(x) = (\sin x)^{\ln x} = e^{(\ln(x))(\ln(\sin x))}$$
or you can use logarithmic differentiation.
If $y = (\sin x)^{\ln x}$, then taking logarithms on both sides we get
$$\ln y = \ln\left((\sin x)^{\ln x}\right) = (\ln x)\ln(\sin x).$$
Now using implicit differentiation we have:
$$\begin{align*}
\frac{d}{dx}\ln y &= \frac{d}{dx}\left( (\ln x)\ln(\sin x)\right)\\
\frac{1}{y}\frac{dy}{dx} &= \left(\ln x\right)'\ln(\sin x) + (\ln x)\left(\ln (\sin x)\right)'\\
\frac{y'}{y} &= \frac{1}{x}\ln(\sin x) + (\ln x)\left(\frac{1}{\sin x}(\sin x)'\right)\\
\frac{y'}{y} &= \frac{\ln\sin x}{x} + \frac{(\ln x)\cos x}{\sin x}\\
\frac{y'}{y} &= \frac{\ln \sin x}{x} + \ln x\cot x\\
y' &= y\left(\frac{\ln \sin x}{x} + \ln x\cot x\right)\\
y' &= (\sin x)^{\ln x}\left(\frac{\ln \sin x}{x} + \ln x\cot x\right).
\end{align*}$$
