Using chain rule take the derivative of $x^{\sin(x)}$

Question

How do I use specifically the chain rule for $x^{\sin(x)}$?

Answer 1

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.)

Answer 2

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.

Answer 3

The power rule of differentiation $$ (x^u)'=ux^{u-1} $$ is only valid for a constant $u\in\mathbb{R}$. It is not valid if $u$ is a function of $x$ itself.