Answer
$$\left( {\text{a}} \right)dy = - \frac{1}{{{x^2}}}dx{\text{ and }}\left( {\text{b}} \right)\,dy = 5{\sec ^2}xdx$$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)y = \frac{1}{x} \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \,\,\,\frac{{dy}}{{dx}} = - \frac{1}{{{x^2}}} \cr
& {\text{Then}} \cr
& \,\,dy = - \frac{1}{{{x^2}}}dx \cr
& \cr
& \left( {\text{b}} \right)y = 5\tan x \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \,\,\,\frac{{dy}}{{dx}} = 5{\sec ^2}x \cr
& {\text{Then}} \cr
& \,\,dy = 5{\sec ^2}xdx \cr
& \cr
& \left( {\text{a}} \right)dy = - \frac{1}{{{x^2}}}dx{\text{ and }}\left( {\text{b}} \right)\,dy = 5{\sec ^2}xdx \cr} $$
