Answer:
[tex]\boxed{(x^{\sin x})' =x^{\sin x}\left(\cos x \log x + \dfrac{\sin x}{x}\right)}.[/tex]
Step-by-step explanation:
First, we should rewrite the expression as:
[tex]x^{\sin x} = \exp[\log(x^{\sin x})] = \exp(\sin x \log x) = e^{\sin x \log x}.[/tex]
We now use the chain rule to get:
[tex](e^{\sin x \log x})' = e^{\sin x \log x}(\sin x \log x)' = x^{\sin x}(\sin x \log x)'.[/tex]
The derivative of the product is given by:
[tex](\sin x \log x)' = (\sin x)' \log x + \sin x (\log x)' = \cos x \log x + \dfrac{1}{x}\sin x.[/tex]
Substituiting, we get:
[tex]x^{\sin x}(\sin x \log x)' = x^{\sin x}\left(\cos x \log x + \dfrac{\sin x}{x}\right).[/tex]
So the answer is:
[tex]\boxed{(x^{\sin x})' =x^{\sin x}\left(\cos x \log x + \dfrac{\sin x}{x}\right)}.[/tex]
