Using chain rule take the derivative of $x^{\sin(x)}$ How do I use specifically the chain rule for $x^{\sin(x)}$?
In your application of the chain rule you have defined 

$$ v(u) = x^u, \quad \text{and} \quad u(x)=\sin(x), $$

and noticed that it follows that

$$ v(u(x)) = x^{\sin(x)}, $$

in order to apply the rule to $v(u(x))$.

The problem is that $v(u)$ as you defined it also depends on $x$, which is the argument of the "inside function", $u(x)$. This invalidates the application of the rule.
Treating the exponent as a constant, the derivative is $\sin(x)\cdot x^{\sin(x)-1}$, treating the basis as a constant the derivative is $x^{\sin(x)}\cdot\ln(x)\cdot\cos(x)$. Adding both gives the derivative of $x^{\sin(x)}$.  (This is not a coincidence.)