Question

c. $h(x) = \int_{1}^{\sqrt{x}} \frac{z^2}{z^4 + 1} dz$

Explanation:

Step1: Recall Leibniz Rule

The Leibniz integral rule for differentiation under the integral sign states that if \( h(x)=\int_{a(x)}^{b(x)} f(z) dz \), then \( h'(x)=f(b(x))\cdot b'(x)-f(a(x))\cdot a'(x) \). Here, \( a(x) = 1\) (a constant), \( b(x)=\sqrt{x}\), and \( f(z)=\frac{z^{2}}{z^{4}+1} \).

Step2: Find derivatives of bounds

First, find \( b'(x) \). Since \( b(x)=\sqrt{x}=x^{\frac{1}{2}} \), using the power rule \( \frac{d}{dx}(x^{n})=nx^{n - 1}\), we get \( b'(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}} \). The derivative of \( a(x) = 1\) is \( a'(x)=0 \).

Step3: Apply Leibniz Rule

Now, substitute into the Leibniz rule. Since \( a'(x) = 0\), the second term vanishes. We have \( f(b(x))=f(\sqrt{x})=\frac{(\sqrt{x})^{2}}{(\sqrt{x})^{4}+1}=\frac{x}{x^{2}+1} \). Then \( h'(x)=f(\sqrt{x})\cdot b'(x)-f(1)\cdot0=\frac{x}{x^{2}+1}\cdot\frac{1}{2\sqrt{x}} \).

Step4: Simplify the expression

Simplify \( \frac{x}{x^{2}+1}\cdot\frac{1}{2\sqrt{x}} \). We know that \( \frac{x}{\sqrt{x}}=\sqrt{x} \), so the expression becomes \( \frac{\sqrt{x}}{2(x^{2}+1)} \).

Answer:

The derivative of \( h(x) \) is \( \boldsymbol{\frac{\sqrt{x}}{2(x^{2}+1)}} \) (if the question was to find the derivative; if it was another operation, adjust accordingly. Assuming differentiation here as it's a common problem with such integrals).