TITLE: How to find the derivative of a function defined by an integral? Namely, $f(y)=\int_0^{y^2} e^{-x^2y^2}dx$
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: 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:
$$
f'(y) = -\frac{1}{y}f(y) + \frac 1y\left(e^{-(y^3)^2}\cdot 3y^2\right) 
= -\frac{f(y)}{y} +3y e^{-y^6}
$$