Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.
$$h ( x ) = \int _ { 1 } ^ { \sqrt { x } } \frac { 5 z ^ { 2 } } { z ^ { 4 } + 1 } d z$$
A) $\frac { 5 \sqrt { x } } { x ^ { 2 } + 1 }$
B) $\frac { \sqrt { x ^ { 2 } + 1 } } { 5 }$
C)
$$\frac { 5 \sqrt { x } } { 2 \left( x ^ { 2 } + 1 \right) }$$
D) $\frac { \sqrt { x + 5 } } { x ^ { 2 } + 5 }$
E) none of these

Question

Quizplus · Accepted Answer

According to Part 1 of the Fundamental Theorem of Calculus, if $f(x) = \int_{a(x)}^{b(x)} g(t) \, dt$, then $f'(x) = g(b(x))b'(x) - g(a(x))a'(x)$. Applying this to our problem, we have $h(x) = \int_1^{\sqrt{x}} \frac{5z^2}{z^4 + 1} \, dz$, so $h'(x) = \frac{5\sqrt{x}}{\sqrt{x^4 + 1}} \cdot \frac{1}{2\sqrt{x}} - \frac{5}{1^4 + 1} \cdot 0 = \frac{5}{2(x^2 + 1)}$. Thus, the answer is $\boxed{	extbf{(C)}}$.

Use Part 1 of the Fundamental Theorem of Calculus to Find