How to differentiate $f(x)=\int_{0}^{\sqrt{x}} e^{-\frac{u^{2}}{x}} d u$ to find $f(x)$ I am asked the function $f(x)$ and given that $f(0)=0$ and
$f(x)=\int_{0}^{\sqrt{x}} e^{-\frac{u^{2}}{x}} d u$
How can I proceed differentiate this function?
I don't know how to apply Newton-Leibniz theorem here.
 A: I suggest writting this function like this:
$$
H(x,y)=\int_0^y e^{-\frac{u^2}{x}}du
$$
So you derivate respect $x$ and $y$. For $y$ use the Fundamental Theorem of Calculus. And for $x$ the derivation under integral symbol.
$$
\dfrac{\partial H(x,y)}{\partial x}=\dfrac{1}{x^2}\int_0^yu^2e^{-\frac{u^2}{x}}du
$$
$$
\dfrac{\partial H(x,y)}{\partial y}=e^{-\frac{y^2}{x}}
$$
Now you do $y=\sqrt{x}$, so $f'(x)=\dfrac{\partial H(x,\sqrt{x})}{\partial x}$.
And you have that
$$
f'(x)=\dfrac{\partial H(x,\sqrt{x})}{\partial x}+\dfrac{\partial H(x,\sqrt{x})}{\partial y}
$$
For that
$$
f'(x)=\dfrac{1}{x^2}\int_0^{\sqrt{x}}u^2e^{-\frac{u^2}{x}}du+e^{-\frac{\sqrt{x}^2}{x}}
$$
$$
f'(x)=\dfrac{1}{x^2}\int_0^{\sqrt{x}}u^2e^{-\frac{u^2}{x}}du+1
$$
You can use the integration by parts(with the functions $u$ and $\dfrac{u}{x}e^{-\frac{u^2}{x}}$) and the integral is
$$
\dfrac{1}{x^2}\int_0^{\sqrt{x}}u^2e^{-\frac{u^2}{x}}du=-\dfrac{1}{2x}\sqrt{x}+\dfrac{1}{2x}\int_0^{\sqrt{x}}e^{-\frac{u^2}{x}}du
$$
So you get the differential equation:
$$
f'(x)=\dfrac{1}{2x}f(x)-\dfrac{1}{2x}\sqrt{x}+1
$$
With the initial condition $f(0)=0$.
