Answer
$\frac{{dy}}{{dx}} = \frac{{\sqrt {1 + x} }}{{{x^2}}}$
Work Step by Step
$$\eqalign{
& y = \int_{1/x}^4 {\sqrt {1 + \frac{1}{t}} } dt \cr
& {\text{Use the integral property }}\int_a^b {f\left( x \right)} dx = - \int_b^a {f\left( x \right)} dx{\text{ }}\left( {{\text{page 391}}} \right) \cr
& y = - \int_4^{1/x} {\sqrt {1 + \frac{1}{t}} } dt \cr
& {\text{Let }}u = \frac{1}{x},{\text{ we can use Part 1 of the Fundamental Theorem }} \cr
& {\text{of calculus with the chain rule as follows}} \cr
& \frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}} \cr
& {\text{Substituting }}y = - \int_4^{1/x} {\sqrt {1 + \frac{1}{t}} } dt \cr
& \frac{{dy}}{{dx}} = \frac{d}{{du}}\left[ { - \int_4^{1/x} {\sqrt {1 + \frac{1}{t}} } dt} \right]\frac{{du}}{{dx}} \cr
& \frac{{dy}}{{dx}} = - \frac{d}{{du}}\left[ {\int_4^{1/x} {\sqrt {1 + \frac{1}{t}} } dt} \right]\frac{{du}}{{dx}} \cr
& {\text{Substitute }}u{\text{ for }}\frac{1}{x} \cr
& \frac{{dy}}{{dx}} = - \frac{d}{{du}}\left[ {\int_4^u {\sqrt {1 + \frac{1}{t}} } dt} \right]\frac{{du}}{{dx}} \cr
& {\text{Use Part 1 of the Fundamental Theorem of calculus}} \cr
& \frac{{dy}}{{dx}} = - \sqrt {1 + \frac{1}{u}} \frac{d}{{dx}}\left[ {\frac{1}{x}} \right] \cr
& \frac{{dy}}{{dx}} = - \sqrt {1 + \frac{1}{u}} \left( { - \frac{1}{{{x^2}}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{x^2}}}\sqrt {1 + \frac{1}{u}} \cr
& {\text{Write in terms of }}x{\text{ substitute }}\frac{1}{x}{\text{ for }}u \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{x^2}}}\sqrt {1 + \frac{1}{{1/x}}} \cr
& \frac{{dy}}{{dx}} = \frac{{\sqrt {1 + x} }}{{{x^2}}} \cr} $$
