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

Question

Quizplus · Accepted Answer

We use Part 1 of the Fundamental Theorem of Calculus, which states that if $h(x)=\int_{a(x)}^{b(x)}f(t)dt$, then $h'(x)=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x)$. In this case, $h(x)=\int_{1}^{\sqrt{x}} \frac{z^{2}}{z^{4}+1} d z$, so $h'(x)=\frac{\sqrt{x}^{2}}{\sqrt{x}^{4}+1}\cdot\frac{1}{2\sqrt{x}}- \frac{1}{1^{4}+1}\cdot0$. Simplifying the first term gives us $\frac{1}{2(x+1)}$, so $h'(x)=\frac{\sqrt{x}}{2(x+1)}=\frac{\sqrt{x}}{2(x^{2}+1)}$. Therefore, the answer is E.

Use Part 1 of the Fundamental Theorem of Calculus to Find