Question

7-18 use part 1 of the fundamental theorem of calculus to find the derivative of the function.

Answer 1

$\displaystyle h(x)=\int\limits_{1}^{\sqrt{x}}~\cfrac{z^2}{z^4+1}dz~\hspace{10em}\cfrac{dh}{du}\cdot \stackrel{chain~rule}{\cfrac{du}{dx}\implies \cfrac{dh}{dx}} \\\\[-0.35em] ~\dotfill\\\\ u=\sqrt{x}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}} \\\\[-0.35em] ~\dotfill$

$\cfrac{dh}{dx}\implies \displaystyle \cfrac{d}{du}\left[ \int\limits_{1}^{u}~\cfrac{z^2}{z^4+1}dz \right]\cdot \cfrac{1}{2\sqrt{x}}\implies \left[ \cfrac{u^2}{u^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}} \\\\\\ \stackrel{substituting~back}{\left[ \cfrac{(\sqrt{x})^2}{(\sqrt{x})^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}}}\implies \cfrac{x}{x^2+1}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{\sqrt{x}}{2x^2+2}$