Q:

Taking the derivative of an integral using chain rule

$2 \frac d {dy} (\int_0^{\sqrt y}3x^2 dx) $
I know that this gives you $3y^{\frac 1 2}$ as a result, if done step by step, but I've been told I can use chain rule to to do it in a single step. I've been staring at it for hours and I just don't see it. Can someone kindly spell it out for me?

A:

Suppose you wanted to calculate
$$
\frac{d}{dy}\int_0^y 3t^2dt
$$
Then you would obtain, by the Fundamental Theorem of Calculus,
$$
3y^2
$$
But, you want to calculate
$$
\frac{d}{dy}\int_0^{f(y)} 3t^2dt
$$
where $f(y):=\sqrt{y}$. So, by the Chain Rule, this is
$$
\left(\frac{d}{ds}\int_0^{s} 3t^2dt\bigg|_{s=f(y)}\right)f'(y)
$$
which is
$$
3\left(\sqrt{y}\right)^2\cdot\frac{1}{2\sqrt{y}}
$$
Simplifying and multiplying by $2$ we get $3y^{\frac{1}{2}}$.

