Find the derivative of $f(x)=\int_{\ln x}^{\sin x} \cos u du$

We should use the derivative formula 
\[f(x)=\int_{g(x)}^{h(x)} k(u)du    \Rightarrow    f'(x)=h'(x)k(h(x))-g'(x)k(g(x)).\]
In this exmple $k(u)=\cos u$, $h(x)=\sin x$, and $g(x)=\ln x$. Hence
\[f'(x)=(\sin x)' \cos (\sin x)-(\ln x)' \cos (\ln x)=\cos x \cos (\sin x)-\frac{1}{x}\cos (\ln x).\]