The value of $f'(\sqrt{\pi})+g'(\sqrt{\pi}).$ Let $f(x)=(\int_{0}^{x}e^{-t^{2}}dt)^{2}$ and $g(x)=\int_{0}^{1}\frac{e^{-x^{2(1+t^{2})}}}{1+t^{2}}dt.$ Then what is the value of $f'(\sqrt{\pi})+g'(\sqrt{\pi})?$ According to me $g'(\sqrt{\pi})$ is equal to zero. But i dont't know how to find $f'(\sqrt{\pi})$. I am thinking it by fundamental theorem of calculus. Please suggest me to find its value. Thanks.


I dont't know how to find $f'(\sqrt{\pi}).$

By applying the chain rule to
$$
f(x)=\left(\int_{0}^{x}e^{-t^{2}}dt\right)^{2}
$$ one gets
$$
f'(x)=2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt
$$ putting $x=\sqrt{\pi}$ gives a result in terms of the error function.
