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?
 A: 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.
A: Five different answers, and all of them using exponentials and logarithms? While logs are indeed convenient, it's certainly possible to solve this problem without what one might consider a detour into logarithm-land.
We know that $(x^n)'=nx^{n-1}$ and $(a^x)'=a^x\log a$, which means that
$$(y^z)'=zy^{z-1}y'$$
when $z$ is constant, and
$$(y^z)'=y^{z}\log y\cdot z'$$
when $y$ is constant. Therefore, when both $y$ and $z$ vary, we have
$$(y^z)'=zy^{z-1}y'+y^{z}\log y\cdot z'.$$
Substitute $y = \sin x$ and $z = \log x$, and you're done.
A: Since
$$
f(x)=\exp[\ln(x)\ln(\sin x)],
$$it
 follows that
$$
f'(x)=[\ln(x)\ln(\sin x)]'f(x)=\left(\frac{\ln(\sin x)}{x}+\cot(x)\ln x\right)(\sin x)^{\ln x}.
$$
A: When evaluating these type of derivatives, the best strategy, in my opinion, is to consider it as an exponential function:
$$f(x)= \sin x^{\log x}$$
$$f(x)=\exp \left[\log (\sin x^{\log x})\right ]$$
$$f(x)=\exp \left[\log x \cdot \log(\sin x)\right]$$
Then we use the chain rule
$$f'(x)=\exp \left[\log x \log(\sin x)\right] \cdot\left(\log x \cdot\log(\sin x)\right)'$$
And that derivative is found by a simple use of the product rule and another chain rule.
A: $$
\begin{align}
f(x) & = (\sin x)^{\ln x} \\[10pt]
\ln f(x) & = \ln \Big((\sin x)^{\ln x} \Big) \\[10pt]
\ln f(x) & = (\ln x) \Big(\ln\sin x\Big) \\[10pt]
\frac{1}{f(x)}\cdot f'(x) & = \text{an expression the comes from the product rule and the chain rule.} \\[10pt]
f'(x) & = f(x)\cdot\Big(\text{an expression the comes from the product rule and the chain rule}\Big).
\end{align}
$$
A: We will need restrictions on $x$, since we cannot allow $\sin x$ to be negative.
From $y=(\sin x)^{\ln x}$, my inclination is to take the logarithm, getting
$$\ln y=\ln(x) \ln(\sin x).$$
and then differentiate implicitly.
In effect, this is the same as the other suggestions. But with a more complicated product of functions, the technique, called logarithmic differentiation, can be quite useful.
A: Our original is:
$$y = (\sin x)^{\ln x} \tag{1}$$
Using logarithmic differentiation, which is helpful for complicated, messy functions such as this, we get:
$$\ln y = \ln (\sin x)^{\ln x}$$
Recall that $\ln a^b = b\ln a$, so we have:
$$\ln y = \ln x \cdot \ln (\sin x)$$
Taking the derivative of both sides gets us:
$$\frac{1}{y} \cdot y' = \ln x \cdot \frac{1}{\sin x} \cdot \cos (x) + \ln(\sin x) \cdot \frac{1}{x} \ \ \ \ \ \text{Product Rule}$$
Cleaning up a bit, we have:
$$\frac{y'}{y} = \ln x \cdot \cot (x) + \frac{\ln{(\sin x)}}{x}$$
Now multiply through by $y$ which is our original, equation $(1)$, and you get:
$$y' = (\sin x)^{\ln x} \left( \ln x \cdot \cot (x) + \frac{\ln{(\sin x)}}{x} \right) $$
