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