## Calculus: Early Transcendentals 8th Edition

$$h'(x) = \frac{\sqrt{x}}{2(x^2 + 1)}$$
$$h(x) = \int\limits^{\sqrt{x}}_1 {\frac{z^2}{z^4 + 1}}dz$$ Using FTC 1, plug the upper bound into z and multiply by the derivative of the upper bound (chain rule). $$h'(x) = \frac{(\sqrt{x})^2}{(\sqrt{x})^4 + 1} \times (\sqrt x)'$$ Simplify. $$h'(x) = \frac{x}{x^2 + 1} \times (\frac{1}{2\sqrt x})$$ $$h'(x) = \frac{\sqrt{x}}{2(x^2 + 1)}$$