If $f(x) = \int_{\cos x}^0 \tan(t)\mathrm dt$, what is $f'(x)$? If $f(x) = \int_{\cos x}^0 \tan(t)\mathrm dt$, what is $f'(x)$?
This is basically the First Fundamental Theorem of Calculus.

\begin{align}
f(x)&amp;=\int^0_{\cos(x)}\tan(t)dt\\
&amp;=-\int_0^{\cos(x)}\tan(t)dt\\
f'(x) &amp; = -\dfrac{d}{d\cos(x)}\int_0^{\cos(x)}\tan(t)dt\ \cdot\dfrac{d\cos(x)}{dx}\\
&amp;= \sin(x)\cdot\tan(\cos(x))
\end{align}
Use the rule $$\frac{d}{dx}(\int_a^x{f(t)dt})=f(x)$$
And the composition rule.

Let $f(x)=\int_{\cos x}^0{\tan t dt}=-h(\cos x)$ where $h(x)=\int_0^{x}{\tan t dt}$

So, $f'(x)=\frac{d}{dx}f(x)=-h'(\cos x).(\cos x)'=\sin x.\tan(\cos x)$