Derivative of a sum involving ln(cos x) and arctan x

For the first term, apply the chain rule to $\ln(\cos(x))$: with outer function $\ln(u)$ and inner function $u = \cos(x)$, we get $\frac{d}{dx}[\ln(\cos(x))] = \frac{-\sin(x)}{\cos(x)}$, which simplifies to $-\tan(x)$.

For the second term, use the known derivative formula $\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}$. Adding the two results gives $f'(x) = -\tan(x) + \dfrac{1}{1+x^2}$.