Answer
$F^{\prime}(x)=2x^{-5}$
Work Step by Step
$F(x)=\displaystyle \int_{2}^{x^{2}}\frac{1}{t^{3}}dt=\int_{2}^{x^{2}}t^{-3}dt$
$=[\displaystyle \frac{1}{-2}\cdot t^{-2}]_{2}^{x^{2}}$
$=-\displaystyle \frac{1}{2}x^{-4}+\frac{1}{2\cdot 2^{4}}$
$=-\displaystyle \frac{1}{2}x^{-4}+\frac{1}{8}$
$F(x)=-\displaystyle \frac{1}{2}x^{-4}+\frac{1}{8}$
$F^{\prime}(x)=-\displaystyle \frac{1}{2}(-4x^{-5})=2x^{-5}$
