How to take the derivative of an integral? $$\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac 2{\sqrt{2\pi}}\int_0^{\sqrt y} \exp\left(-{\frac{x^2}{2}}\right) \,\mathrm{d}x\right).$$
I try to integrate first and then do the differentiation but it's not easy. I want to know other way to do it. Thank you.
 A: We apply the Leibniz integral rule
This gives us,
$$\frac{2}{\sqrt{2\pi}}\left( e^{-(\sqrt{y})^2}\cdot\left(\frac{d}{dy}\sqrt{y}\right) - e^{-0^2}\cdot\left(\frac{d}{dy} 0\right)\right),\\
=\frac{2}{\sqrt{2\pi}}e^{-y}\frac{1}{2\sqrt{y}},\\
=\frac{1}{\sqrt{2\pi}}e^{-y}\frac{1}{\sqrt{y}}$$
A: the first derivative is :
$$ (\frac{2}{{\sqrt{2\pi}}})\frac{1}{2{\sqrt{y}}}e^{-x^{2}/2}$$
A: HINT
Note that in general
$$f(t)=\int_{a(t)}^{b(t)}g(u) du\implies f'(t)=g(b(t))\cdot b'(t)-g(a(t))\cdot a'(t)$$
thus
$$\frac d{dy}\left(\frac 2{\sqrt{2\pi}}
\int_0^{\sqrt y} e^{\frac {-x^2}{2}}dx\right)=\frac 2{\sqrt{2\pi}}(\sqrt y)'e^{\frac {-y^2}{2}})=\frac 1{\sqrt{2\pi}}\,\frac{1}{\sqrt y}\,e^{\frac {-y}{2}}$$
A: If you define $F(y)=\int_0^ye^{-\frac{x^2}2}~dx$ then by fundamental lemma of calculus $F$ is differentiable with $F'(y)=e^{-\frac{y^2}2}$. 
We use $\frac{d}{dy}\sqrt{y}=\frac1{2\sqrt{y}}$ and the chain rule and get
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac 2{\sqrt{2\pi}}\int_0^{\sqrt y} \exp\left(-{\frac{x^2}{2}}\right) \,\mathrm{d}x\right)&=\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac 2{\sqrt{2\pi}}F(\sqrt{y}) \right)\\
&=\frac2{\sqrt{2\pi}}F'(\sqrt{y})\cdot\frac1{2\sqrt{y}}\\
&=\frac1{\sqrt{2\pi y}}e^{-\frac{y}2}.
\end{align*}
