Answer
$$ f'(x)= x^{\cos x}(-\sin x\ln x+\frac{\cos x}{x}).$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(\ln x)'=\dfrac{1}{x}$
Recall that $(\sin x)'=\cos x$.
Recall the product rule: $(uv)'=u'v+uv'$
Using the method from Example 9, we get:
$$ f(x)=(e^{\ln x})^{\cos x}=e^{\cos x\ln x}.$$
Now taking the derivative, we get
$$ f'(x)= e^{\cos x\ln x}(\cos x\ln x)'=e^{\cos x\ln x}(-\sin x \ln x+\cos x /x) =e^{\cos x\ln x}(-\sin x \ln x+\cos x/x)=x^{\cos x}(-\sin x\ln x+\frac{\cos x}{x}).$$
