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So I need to calculate $f'(\sqrt{\pi})+g'(\sqrt{\pi})$ of the integrals $f(x)=(\int_{0}^{x}e^{-t^2}dt)^2$ and $g(x)=\int_{0}^{1}\frac{e^{-x^2(1+t^2)}}{1+t^2}dt$

First I used differentiation under the integral sign rule and try to solve it,

for the first integral , I got $f'(x)=2 (\int_{0}^{x}e^{-t^2}dt) e^{-x}$, then $f'(\sqrt\pi)=2 (\int_{0}^{\sqrt\pi}e^{-t^2}dt) e^{-\sqrt\pi}$. Now I don't know how to solve the integral from $0$ to $\sqrt\pi$. I am also unable to solve the next integral as well, as same integral occurs after differentiantion. Please help me to solve this.

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$$f(x)=\left(\int_{0}^{x}e^{-t^{2}}dt\right)^{2} \implies f'(x)=2e^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$ \begin{align*} g'(x)=\int_{0}^{1}\frac{\frac{\partial}{\partial x}\left( e^{-x^{2}(1+t^{2})}\right)}{1+t^{2}}dt&=-\int_{0}^{1}2xe^{-x^{2}(1+t^{2})}dt\\ &=-2xe^{-x^2}\int_{0}^{1}e^{-x^2t^2}dt\\ &=-2e^{-x^{2}}\int_{0}^{x}e^{-z^{2}}dz\qquad \text{substituting $xt=z$} \end{align*} Hence $f'(x)+g'(x)=0$ for all $x$.

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your first integral is wrong,first integral is for the fundamental theorem of calculus and the chain rule, equal to: $$f(x)=\left(\int_{0}^{x}e^{-t^{2}}dt\right)^{2} =>f'(x)=-2xe^{-x^{2}}\int_{0}^{x}e^{-t^{2}}dt$$ this integral don't have analytic solution for x finite, just numerical, then the result is in term of the integral. The second function: $$g'(x)=\frac{d}{dx}\int_{0}^{1}\frac{e^{-x^{2}(1+t^{2})}}{1+t^{2}}dt$$ using the Liebniz rule, the differential go to inside the integral, because this not depends of $x$: $$g'(x)=\int_{0}^{1}\frac{\frac{d}{dx}\left( e^{-x^{2}(1+t^{2})}\right)}{1+t^{2}}dt=-\int_{0}^{1}2xe^{-x^{2}(1+t^{2})}dt$$ finally the solution is: $$f'(\sqrt{\pi})+g'(\sqrt{\pi})=-2\sqrt{\pi} e^{-\pi}\int_{0}^{\sqrt{\pi}}e^{-t^{2}}dt-2\sqrt{\pi}\int_{0}^{1}e^{-\pi(1+t^{2})}dt=$$ $$=-4\sqrt{\pi}e^{-\pi}\left(\int_{0}^{1}e^{-\pi t^{2}}dt+\int_{0}^{\sqrt{\pi}} e^{-t^{2}}dt\right)$$

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