Fundamental theorem of calculus tricky question Let 
$$F(x) = \int_{0}^{\sqrt{x}}(\sqrt{x}-t^2) f(t) dt$$ 
and
$$G(x) = \int_{0}^{\sqrt{x}}f(t) dt$$ 
Find a formula for $F'(x)$ in terms of $f$ and $G$.
I know that I have to use fundamental theorem of calculus but just do not know how to handle the int f(t)(t) dt part. 
 A: You can rewrite:
$$F(x) = \sqrt{x}\int_{0}^{\sqrt{x}}f(t)dt - \int_{0}^{\sqrt{x}}t^2f(t) dt$$
$$F(x) = \sqrt{x}G(x) - \int_{0}^{\sqrt{x}}t^2f(t) dt$$
Do a variable substitution for the integral. Let $a=t^2$. 
$da = 2tdt$
$da = 2\sqrt{a}dt$
$dt = \frac{da}{2\sqrt{a}}$
$$F(x) = \sqrt{x}G(x) - 0.5\int_{0}^{x}\sqrt{a}f(\sqrt{a}) da$$
Use fundamental theorem of calculus for the integral.
$$F'(x) = \frac{0.5G(x)}{\sqrt{x}} + \sqrt{x}G'(x) - 0.5\sqrt{x}f(\sqrt{x})$$
EDIT:
I think maybe the question wants no G'(x) in the final answer. We can rewrite G'(x) in terms of f(x)
$$G(x) = \int_{0}^{\sqrt{x}}f(t)dt$$
Do the same $a=t^2$ substitution
$$G(x) = \int_{0}^{x}\frac{f(\sqrt{a})}{2\sqrt{a}}da$$
$$G'(x) = \frac{f(\sqrt{x})}{2\sqrt{x}}$$
Substitute into our formula for $F'(x)$
$$F'(x) = \frac{0.5G(x)}{\sqrt{x}} + 0.5f(\sqrt{x})(1-\sqrt{x})$$
A: Write $F(x)$ as $$\sqrt x\int_0^{\sqrt x}f(t)\mathrm dt-\int_0^{\sqrt x}t^2f(t)\mathrm dt=\sqrt xG(x)-\int_0^{\sqrt x}t^2f(t)\mathrm dt.$$ Then differentiating the first term using the product rule gives $$\frac12\frac{1}{\sqrt x}G(x)+\sqrt x G'(x).$$ Differentiating the second part gives $$-\frac12\frac{1}{\sqrt x}(\sqrt x)^2f(\sqrt x).$$ Note that $$G'(x)=\frac12\frac{1}{\sqrt x}f(\sqrt x).$$
