How to find the derivative of a function defined by an integral? Namely, $f(y)=\int_0^{y^2} e^{-x^2y^2}dx$ Find at each point of its domain the derivative of the function $f: \mathbb{R} \rightarrow \mathbb{R}$ $$f(y)=\int_0^{y^2} e^{-x^2y^2}dx$$
$$$$
Is the domain of the function $\mathbb{R}$ because of the exponential?
Do I have to set $f(y)=h(y^2)$ and then use 
$$\frac{d h(y^2)}{dy}=\frac{d h(y^2)}{dy^2} \cdot \frac{dy^2}{dy}$$
Or is there an other way to find the derivative?
 A: Substitute $u=xy \Rightarrow du=ydx \Rightarrow dx=\frac{du}y$.
$$
f(y)=\int_0^{y^2} e^{-x^2y^2}dx = \int_0^{y^3} e^{-u^2}\frac{du}y = \frac 1y \int_0^{y^3} e^{-u^2}du
$$
Now suppose we have the anti-derivative:
$$F(u):=\int_0^u e^{-u^2}du$$
Then:
$$F'(u)=e^{-u^2} \tag{1}$$
and:
$$
f(y)=\frac 1y \int_0^{y^3} e^{-u^2}du =\frac 1y \left(F(y^3)-F(0)\right)\tag{2}
$$
Now we are ready to take the derivative:
$$
f'(y) = -\frac{1}{y^2}\left(F(y^3)-F(0)\right) + \frac 1y\left(F'(y^3)\cdot 3y^2\right)
$$
And when we substitute $(1)$ and $(2)$ back, we get:
$$
f'(y) = -\frac{1}{y}f(y) + \frac 1y\left(e^{-(y^3)^2}\cdot 3y^2\right) 
= -\frac{f(y)}{y} +3y e^{-y^6}
$$
A: Hint:Use leibnitz integral rule
$\frac{d}{dy}\ (\int_{a(y)}^{b(y)}f(x,y)dx)=f(b(y),y)b'(y)-f(a(y),y)a'(y)+\int_{a(y)}^{b(y)}f_y(x,y)dx$
A: The domain of $f$ is set of points $y$ such that $\displaystyle \int \limits_0^{y^2}e^{-x^2y^2}\mathrm dx$ exists. 
For any fixed $y$, the function $x\mapsto e^{-x^2y^2}$ is integrable is any interval of the form $[a,b]$ - because it is continuous there - (in fact it is integrable in $\mathbb R$, but that doesn't matter for the purpose of this problem), in particular it is integrable in $[0,b]$ for all $b\in \mathbb R$ and particularizing further, it is integrable in $[0,y^2]$, that is, $\displaystyle \int \limits_0^{y^2}e^{-x^2y^2}\mathrm dx$ makes sense.
Therefore $\text{dom}(f)=\mathbb R$.
An example in which things don't work so smoothly is, for instance, $\displaystyle g(y)=\int \limits _0^{y}y\sqrt x\,\mathrm dx$. Even though the square root does not affect $y$, the domain of this function can't possibly include negative values because for any fixed $y$, the function $x\mapsto y\sqrt x$ isn't defined for negative values of $x$, you can't integrate over, for example, $[-1,0]$, i.e., $\displaystyle \int \limits _0^{-1}-\sqrt x\,\mathrm dx$ doesn't make sense.
The second part of the problem can be solved with differentiation under the integral sign.
A: Let $z=f(y)$ and $u=y^2$, so $\displaystyle z=\int_{0}^{u} e^{-x^2y^2} dx$.
By the Chain Rule, $\displaystyle\frac{dz}{dy}=\frac{\partial z}{\partial u}\cdot\frac{du}{dy}+\frac{\partial z}{\partial y}\cdot\frac{dy}{dy}$ where
$\displaystyle\frac{\partial z}{\partial u}=e^{-u^2y^2}$ and $\displaystyle\frac{\partial z}{\partial y}=\int_{0}^{u}\frac{\partial}{\partial y}(e^{-x^2y^2})dx$ by the Leibniz Integral Rule.
