What is the error in the solution of the composite derivative?

Question

I have the following derivative to get: $\Large{\frac{d}{dx}x^{\sqrt{\sin(x)}}}$

My attempt was:

Let $u = \sqrt{\sin(x)}$, now i have $\Large{\frac{d}{du}x^u \cdot \frac{d}{dx}u }$

And solving, i got: $\Large{ux^{u-1}\cdot \frac{d}{dx}u}$

Let $k = \sin(x)$ i have $\Large{\frac{d}{dx}u=\frac{d}{dk}\sqrt{k}\cdot \frac{d}{dx}\sin(x) = \frac{\cos(x)}{2\sqrt{k}} }$

Finally i got: $\Large{\frac{d}{dx}x^{\sqrt{\sin(x)}} =u\cdot x^{u-1} \cdot \frac{\cos(x)}{2\sqrt{k}} =\sqrt{\sin(x)} \cdot x^{\sqrt{\sin(x)}-1} \cdot \frac{\cos(x)}{2\sqrt{\sin(x)}}}$.

But the result is not correct and i want to know what is wrong in this development,

Thanks in advance.

Answer 1

Here is the error : $\frac{d}{du}x^u$ is not equals to $ux^{u-1}$. Instead, $\frac{d}{du}x^u =\frac{d}{du} \exp(u \log x)=(\log x)x^u$.

Answer 2

Instead write $x^{\sqrt{\sin x}}=e^{\sqrt{\sin(x)}\ln{x}}$ so:

$\frac{d}{dx}x^{\sqrt{\sin x}}=\frac{d}{dx}e^{\sqrt{\sin(x)}\ln{x}}=\frac{d}{dx}(\sqrt{\sin(x)}\ln x)e^{\sqrt{\sin(x)}\ln x}=x^{\sqrt{\sin(x)}}\left(\frac{\sqrt{\sin(x)}}{x}+\frac{\ln(x)\cos(x)}{2\sqrt{\sin(x)}}\right)$