value of $f'(\sqrt π)+g'(\sqrt π)?$

Question

Let $$f(x)=\left(\int_{0}^{x} e^{-t^2}dt\right)^2$$ and

$$g(x)=\int_{0}^{1} \frac{e^{-x^2(1+t^2)}} {1+t^2} dt$$

Then what is the value of$$f'(\sqrt π)+g'(\sqrt π)?$$ I don't know how to solve this. But I guess in $g(x)$ we need to use gamma function.

Answer 1

One has

$$f’(x)=2e^{-x^2}\int_0^xe^{-t^2}dt $$

And

$$\begin{align}g’(x)&=-\int_0^12{(1+t^2)xe^{-x^2(1+t^2)}\over 1+t^2}dt \\&=-2x\int_0^1e^{-x^2(1+t^2)}dt\end{align}$$

This leads to

$$\begin{align}f’(\sqrt{\pi})+g’(\sqrt{\pi})&=2e^{-\pi}\left(\int_0^{\sqrt{\pi}}e^{-t^2}dt-\sqrt{\pi}\int_0^1e^{-\pi t^2}dt\right)\\&=0\end{align}$$

Answer 2

Hint: $$f'(x)=2\left(\int_{0}^{x} e^{-t^2}dt\right)\cdot \dfrac{d}{dx}\left(\int_{0}^{x} e^{-t^2}dt\right)=2\left(\int_{0}^{x} e^{-t^2}dt\right)\cdot e^{-x^2}$$ and

$$g'(x)=\int_{0}^{1} \frac{\dfrac{d}{dx}\left[e^{-x^2(1+t^2)}\right]} {1+t^2} dt$$