The problem asks to use the Fundamental Theorem of Calculus (FTC) to compute $h'(x)$, where

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

Solution

Using the Second Fundamental Theorem of Calculus, if a function $h(x)$ is defined as:

$h(x) = \int_{a(x)}^{b(x)} f(z) \, dz,$

then its derivative is given by:

$h'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x),$

where $f(z)$ is the integrand, $b(x)$ is the upper limit of integration, and $a(x)$ is the lower limit of integration.

Step 1: Identify $f(z)$, $a(x)$, and $b(x)$

• $f(z) = \frac{z^2}{z^4 + 1}$,
• $a(x) = 1$ (constant),
• $b(x) = \sqrt{x}$.

Step 2: Apply the derivative rule

Since $a(x) = 1$ is constant, $a'(x) = 0$. For $b(x) = \sqrt{x}$, its derivative is:

$b'(x) = \frac{1}{2\sqrt{x}}.$

Using the formula for $h'(x)$:

$h'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x).$

Substituting $f(z) = \frac{z^2}{z^4 + 1}$, $b(x) = \sqrt{x}$, and $a(x) = 1$:

$h'(x) = \frac{(\sqrt{x})^2}{(\sqrt{x})^4 + 1} \cdot \frac{1}{2\sqrt{x}} - \frac{1^2}{1^4 + 1} \cdot 0.$

Step 3: Simplify the expression

• $(\sqrt{x})^2 = x$,
• $(\sqrt{x})^4 = x^2$,
• $a'(x) = 0$, so the second term vanishes.

Thus:

$h'(x) = \frac{x}{x^2 + 1} \cdot \frac{1}{2\sqrt{x}}.$

Simplify further:

$h'(x) = \frac{1}{2} \cdot \frac{\sqrt{x}}{x^2 + 1}.$

Final Answer:

$h'(x) = \frac{\sqrt{x}}{2(x^2 + 1)}.$

Would you like a more detailed explanation or assistance with related questions? Here are some follow-up questions:

1. How does the Fundamental Theorem of Calculus relate to definite integrals?
2. Can $h(x)$ be evaluated directly for specific $x$ values?
3. What if the limits of integration were reversed? How would the derivative change?
4. How does $f(z) = \frac{z^2}{z^4 + 1}$ behave as $z \to \infty$?
5. What is the geometric interpretation of $h(x)$?

Tip: Always double-check your differentiation, especially when chain rule components like $b'(x)$ are involved!