Taking the derivative of an integral using chain rule

Question

$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?

Answer 1

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 $3y^{\frac{1}{2}}$.

Answer 2

Chain rule says the following:

$$\frac{d}{dy} \int_{h(y)}^{g(y)} f(x)dx = f(g(y))\cdot g'(y) - f(h(y))\cdot h'(y).$$

Identify your $f(x),\,h(y),\,g(y)$ and plug in.