Derivative of $f(x)= (\sin x)^{\ln x}$ I am just  wondering if i went ahead to solve this correctly? 
I am trying to find the derivative of $f(x)= (\sin x)^{\ln x}$
Here is what i got below.
$$f(x)= (\sin x)^{\ln x}$$
$$f'(x)=\ln x(\sin x) \Rightarrow f'(x)=\frac{1}{x}\cdot\sin x + \ln x \cdot \cos x$$
Would that be the correct solution?

It's instructive to look at this particular logarithmic-differentiation situation generally:
$$\begin{align}
y&=u^{v}\\[0.5em]
\implies \qquad \ln y &= v \ln u & \text{take logarithm of both sides}\\[0.5em]
\implies \qquad \frac{y^{\prime}}{y} &= v \cdot \frac{u^{\prime}}{u}+v^{\prime}\ln u & \text{differentiate}\\
\implies \qquad y^{\prime} &= u^{v} \left( v \frac{u^{\prime}}{u} + v' \ln u \right) & \text{multiply through by $y$, which is $u^{v}$} \\
&= v \; u^{v-1} u^{\prime} + u^{v} \ln u \; v^{\prime} & \text{expand}
\end{align}$$
Some (most?) people don't bother with the "expand" step, because right before that point the exercise is over anyway and they just want to move on. (Plus, generally, we like to see things factored.) Even so, look closely at the parts you get when you do bother:
$$\begin{align}
v \; u^{v-1} \; u^{\prime} &\qquad \text{is the result you'd expect from the Power Rule if $v$ were constant.} \\[0.5em]
u^{v} \ln u \; v^{\prime} &\qquad \text{is the result you'd expect from the Exponential Rule if $u$ were constant.}
\end{align}$$
So, there's actually a new Rule here: the Function-to-a-Function Rule is the "sum" of the Power Rule and Exponential Rule!
Knowing FtaF means you can skip the logarithmic differentiation steps. For example, your example:
$$\begin{align}
\left( \left(\sin x\right)^{\ln x} \right)^{\prime} &= \underbrace{\ln x \; \left( \sin x \right)^{\ln x - 1} \cos x}_{\text{Power Rule}} + \underbrace{\left(\sin x\right)^{\ln x} \; \ln \sin x \; \frac{1}{x}}_{\text{Exponential Rule}}
\end{align}$$
As I say, we generally like things factored, so you might want to manipulate the answer thusly,
$$
\left( \left(\sin x\right)^{\ln x} \right)^{\prime} = \left( \sin x \right)^{\ln x} \left( \frac{\ln x \cos x}{\sin x} + \frac{\ln \sin x}{x} \right) = \left( \sin x \right)^{\ln x} \left( \ln x \cot x + \frac{\ln \sin x}{x} \right)
$$
Another example:
$$\begin{align}
\left( \left(\tan x\right)^{\exp x} \right)^{\prime} &= \underbrace{ \exp x \; \left( \tan x \right)^{\exp x-1} \; \sec^2 x}_{\text{Power Rule}} + \underbrace{ \left(\tan x\right)^{\exp x} \ln \tan x \; \exp x}_{\text{Exponential Rule}} \\
&= \exp x \; \left( \tan x \right)^{\exp x} \left( \frac{\sec^2 x}{\tan x} + \ln \tan x \right) \\
&= \exp x \; \left( \tan x \right)^{\exp x} \left( \sec x \; \csc x + \ln \tan x \right)
\end{align}$$
Note. Be careful invoking FtaF in a class --especially on a test-- where the instructor expects (demands) that you go through the log-diff steps every time. (Of course, learning and practicing those steps is worthwhile, because they apply to situations beyond FtaF.) On the other hand, if you explain FtaF to the class, you could be a hero for saving everyone a lot of effort with function-to-a-function derivatives.
