Differentiate ${{x}^{\sin x}},x>0$ with respect to $x$ .

Hint: Start by using $y={{x}^{\sin x}}$ and taking log of both the sides of the equation. Now use the identity that $\log {{a}^{b}}=b\log a$ and differentiate both sides of the equation with respect to x. Use the uv rule of differentiation, i.e., $\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}$ .

Complete step-by-step solution -
Let us start the solution to the above question by letting ${{x}^{\sin x}},x>0$ to be equal to y. So, we can say
$y={{x}^{\sin x}}............(i)$
Now we know, if we take log of both sides of the equation, we know that the equation remains valid. So, we will take log of both sides of the equation. On doing so, we get
$\log y=\log {{x}^{\sin x}}$
Now, we know that $\log {{a}^{b}}=b\log a$ . So, using this identity in our equation, we get
$\log y=\sin x\log x$
Now we will differentiate both sides of the equation with respect to x. On doing so, we get
 $\dfrac{d\left( \log y \right)}{dx}=\dfrac{d\left( \sin x\log x \right)}{dx}$
Using the uv rule of differentiation, i.e., $\dfrac{d\left( uv \right)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}$ and $\dfrac{d\left( \log y \right)}{dx}=\dfrac{1}{y}\dfrac{dy}{dx}$ , we get
$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{\log xd\left( \sin x \right)}{dx}+\dfrac{\sin xd\left( \log x \right)}{dx}$
Now, we know that the derivative of $\sin x\text{ is cosx}$ and $\log x\text{ is }\dfrac{1}{x}$ . So, using this in our equation, we get
$\dfrac{1}{y}\dfrac{dy}{dx}=\log x\times \operatorname{cosx}+\sin x\times \dfrac{1}{x}$
$\Rightarrow \dfrac{dy}{dx}=y\left( \log x\times \operatorname{cosx}+\sin x\times \dfrac{1}{x} \right)$
Now, we will put the value of y from equation (i). On doing so, we get
$\dfrac{dy}{dx}={{x}^{\sin x}}\left( cosx\log x+\dfrac{\sin x}{x} \right)$
Therefore, we can conclude that the derivative of ${{x}^{\sin x}},x>0$ is ${{x}^{\sin x}}\left( cosx\log x+\dfrac{\sin x}{x} \right).$

Note: Be careful with the signs and calculations as in such questions, the possibility of making a mistake is either of the sign or a calculation error. Also, remember that for taking log of both the sides of the equation, both the sides must be positive. In the above question the RHS is positive as it is given that x>0, if it was not mentioned, we could not have taken log of both the sides and solved the equation.