Answer
$h'(x) = \frac{x}{2\sqrt x(x^2+1)}$
Work Step by Step
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $h(x) = \int ^{\sqrt{x}}_{1}\frac{z^2}{z^4+1}dz$
Since the upper bound is a function we need to use the chain rule:
$u = \sqrt{x}$
$u' = \frac{1}{2\sqrt x}$
Since $h(x) = \int^{b}_{a}f(t)dt$
$h'(x) = f(x)$
$h'(u) = \frac{u^2}{u^4+1}u'$
$h'(x) = \frac{\sqrt x^2}{\sqrt x^4+1}\times \frac{1}{2\sqrt x}$
$h'(x) = \frac{x}{x^2+1}\times \frac{1}{2\sqrt x}$
$h'(x) = \frac{x}{2\sqrt x(x^2+1)}$