Answer
$\frac{{dy}}{{dx}} = - \frac{5}{{{x^2} + 1}}$
Work Step by Step
$$\eqalign{
& y = 5\arctan \left( {\frac{1}{x}} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {5\arctan \left( {\frac{1}{x}} \right)} \right] \cr
& \frac{{dy}}{{dx}} = 5\frac{d}{{dx}}\left[ {\arctan \left( {\frac{1}{x}} \right)} \right] \cr
& {\text{Use }}\frac{d}{{dx}}\left[ {\arctan u} \right] = \frac{1}{{1 + {u^2}}}\frac{{du}}{{dx}},{\text{ let }}u = \frac{1}{x} \cr
& \frac{{dy}}{{dx}} = 5\left( {\frac{1}{{1 + {{\left( {1/x} \right)}^2}}}} \right)\frac{d}{{dx}}\left[ {\frac{1}{x}} \right] \cr
& \frac{{dy}}{{dx}} = 5\left( {\frac{1}{{1 + 1/{x^2}}}} \right)\left( { - \frac{1}{{{x^2}}}} \right) \cr
& {\text{Multiply and simplify}} \cr
& \frac{{dy}}{{dx}} = 5\left( {\frac{{{x^2}}}{{{x^2} + 1}}} \right)\left( { - \frac{1}{{{x^2}}}} \right) \cr
& \frac{{dy}}{{dx}} = 5\left( {\frac{1}{{{x^2} + 1}}} \right)\left( { - 1} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{5}{{{x^2} + 1}} \cr} $$
