How do you use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function h(x)= sqrt(3+r^3) dr for a=1, b=x^2?

I wonder if perhaps you mean to ask how to use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function $h \left(x\right) = {\int}_{1}^{{x}^{2}} \sqrt{3 + {r}^{3}} \mathrm{dr}$
$\frac{d}{\mathrm{du}} \left({\int}_{1}^{u} \sqrt{3 + {r}^{3}} \mathrm{dr}\right) = \sqrt{3 + {u}^{3}}$,
but we've been asked for d/(dx)(h(x) so we'll use the chain rule with $u = {x}^{2}$
$\frac{d}{\mathrm{dx}} \left({\int}_{1}^{u} \sqrt{3 + {r}^{3}} \mathrm{dr}\right) = \sqrt{3 + {u}^{3}} \cdot \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) = 2 x \sqrt{3 + {x}^{6}}$,