Finding the derivative of the function $G(t)$

Question

Let $f$ be a differentiable and increasing , meaning that $f'(x)>0$ for all $x$ on the interval $[a,b]$, with $a>0$. Furthermore, $f$ has a differentiable inverse function $f^{-1}$. Let us consider the integrals
$G(t)=$$\int_{f(a)}^{f(t)} \pi((f^{-1}(y))^2-a^2) dy$ and
$H(t)=$$\int_{a}^{t} 2\pi x(f(t)-f(x)) dx$.
Show that $G(t)$ and $H(t)$ have identical derivatives on $[a,b]$. I think we must use Leibniz integral rule here, but how can I apply it to $G(t)$? In case of $H(t)$ it is quite easier than that to apply on $G(t)$. PLease help me to find the derivative of $G(t)$.

Answer 1

$G'(t)=\pi [(f^{-1}f(t))^{2}-a^{2}]f'(t)=\pi [t^{2}-a^{2}]f'(t)$ by chain rule. $H(t)=f(t)\int_a^{t} 2\pi x \, dx-\int_a^{t}2\pi xf(x)\, dx$ so $H'(t)=2\pi tf(t)+\pi {(t^{2}-a^{2})} f'(t)-2\pi f(t)=\pi {(t^{2}-a^{2})} f'(t)$. Hence $H'(t)=G'(t)$.