Find $F'(x)$ of $F(x)=\int_{\pi e^x}^{\cos(x)} \cos (t) dt$

Question

Let $F: \mathbb{R} \to \mathbb{R}$ be a function, $F(x)=\int_{\pi e^x}^{\cos(x)} \cos (t) dt$.

Find $F'(x)$.

I can't get to the correct answer which is $\pi (-e^x) \cos(\pi e^x) - \sin(x) \cos(\cos(x))$.

I used the chain rule, first with the upper bound:

$\cos(\cos(x))(-\sin(x))$.

And then with the lower bound:

$\cos(\pi e^x)(e^x)$.

So $F'(x)=\cos(\cos(x))(-\sin(x))-\cos(\pi e^x)(e^x)$? Am I missing something?

Answer 1

You miss a $\pi$: $$ F'(x)=\cos(\cos(x))(-\sin(x))-\cos(\pi e^x)(e^x)\color{red}{\pi}. $$

Note that $$ \frac{d}{dx}(\pi e^x)=\pi e^x. $$

Answer 2

for some function: $$F(x)=\int_{a(x)}^{b(x)}f(x,t)dt$$ $$F'(x)=f(x,b(x)).\frac{d}{dx}[b(x)]-f(x,a(x)).\frac{d}{dx}[a(x)]+\int_{a(x)}^{b(x)}\frac{\partial}{\partial_x}f(x,t)dt$$ so for our function: $$F(x)=\int_{\pi e^x}^{\cos(x)}\cos(t)dt$$ $$F'(x)=-\cos(\cos(x))\sin(x)-\pi\cos(\pi e^x)e^x+\int_{\pi e^x}^{\cos(x)}0.dt$$ so in your case it all appears correct apart from a missing $\pi$ in the second term, due to the fact that $D[\pi e^x]=\pi e^x$