Question

Use part one of the fundamental theorem of calculus to find the derivative of the function.
h(x) =
√x
z2
z4 + 2
dz

1

Answer 1

To find the derivative of the function defined by the integral

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

we can apply Part 1 of the Fundamental Theorem of Calculus. This theorem states that if \( F(t) \) is an antiderivative of \( f(t) \), then

\[
\frac{d}{dx} \left(\int_{a}^{g(x)} f(t) \, dt \right) = f(g(x)) \cdot g'(x),
\]

where \( g(x) \) is a function of \( x \) and \( g'(x) \) is its derivative.

In our case, we can identify:
- \( f(z) = \frac{z^2}{z^4 + 2} \),
- \( g(x) = \sqrt{x} \),
- the lower limit \( a = 1 \).

Step 1: Calculate \( g'(x) \).

\[
g(x) = \sqrt{x} \Rightarrow g'(x) = \frac{1}{2\sqrt{x}}.
\]

Step 2: Substitute into the derivative formula.

Now we need to compute \( f(g(x)) \):

\[
f(g(x)) = f(\sqrt{x}) = \frac{(\sqrt{x})^2}{(\sqrt{x})^4 + 2} = \frac{x}{x^2 + 2}.
\]

Step 3: Put it all together.

Using the fundamental theorem,

\[
h'(x) = f(g(x)) \cdot g'(x) = \frac{x}{x^2 + 2} \cdot \frac{1}{2\sqrt{x}}.
\]

Step 4: Simplify \( h'(x) \):

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

Thus, the derivative \( h'(x) \) is

\[
\boxed{\frac{\sqrt{x}}{2(x^2 + 2)}}.
\]