Answer
$$y'=(x\cos x\ln x+\sin x)x^{\sin x-1}$$
Work Step by Step
$$y=x^{\sin x}$$
We can also use logarithmic differentiation to solve these types of problems. Take the natural logarithm of both sides: $$\ln y=\ln(x^{\sin x})$$ $$\ln y=\sin x\ln x$$
Now use implicit differentiation to find $y'$ with respect to $x$: $$(\ln y)' =(\sin x\ln x)'$$
$$\frac{y'}{y}=\cos x\ln x+\frac{\sin x}{x}$$
$$\frac{y'}{y}=\frac{x\cos x\ln x+\sin x}{x}$$
Find $y'$: $$y'=\Big(\frac{x\cos x\ln x+\sin x}{x}\Big)y=\frac{(x\cos x\ln x+\sin x)x^{\sin x}}{x}$$
$$y'=(x\cos x\ln x+\sin x)x^{\sin x-1}$$
