Answer
$F^{\prime}(x)=\displaystyle \frac{2}{x}$
Work Step by Step
Use the Second Fundamental Theorem of Calculus (Th 4.11)
$\displaystyle \frac{d}{dx}\left[\int_{a}^{x}f(t)dt\right] =f(x)$
Here, $F(x)$=$\displaystyle \int_{1}^{x^{2}}f(t)dt,\qquad$ f(t) $=\displaystyle \frac{1}{t}, a=1$
With the substitution $u=x^{2}, \displaystyle \frac{du}{dx}=2x,$
we apply the chain rule
$F^{\prime}(x)=\displaystyle \frac{dF}{du}\cdot\frac{du}{dx}$
$=\displaystyle \frac{d}{du}\left[\int_{1}^{u}f(t)\right]\cdot 2x$
$=f(u)\cdot 2x$
$=\displaystyle \frac{1}{u}\cdot 2x$
$=\displaystyle \frac{1}{x^{2}}\cdot 2x$
$=\displaystyle \frac{2}{x}$
