Answer
$$\left( {\text{a}} \right)dy = \frac{{2 - 3x}}{{2\sqrt {1 - x} }}dx{\text{ and }}\left( {\text{b}} \right)\,dy = - \frac{{17}}{{{{\left( {1 + x} \right)}^{18}}}}dx$$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)y = x\sqrt {1 - x} \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \,\,\,\frac{{dy}}{{dx}} = x\left( {\frac{{ - 1}}{{2\sqrt {1 - x} }}} \right) + \sqrt {1 - x} \cr
& \,\,\,\frac{{dy}}{{dx}} = \frac{{ - x + 2\left( {1 - x} \right)}}{{2\sqrt {1 - x} }} \cr
& \,\,\,\frac{{dy}}{{dx}} = \frac{{2 - 3x}}{{2\sqrt {1 - x} }} \cr
& {\text{Then}} \cr
& \,\,dy = \frac{{2 - 3x}}{{2\sqrt {1 - x} }}dx \cr
& \cr
& \left( {\text{b}} \right)y = {\left( {1 + x} \right)^{ - 17}} \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \,\,\,\frac{{dy}}{{dx}} = - 17{\left( {1 + x} \right)^{ - 18}}\left( 1 \right) \cr
& {\text{Then}} \cr
& \,\,dy = - \frac{{17}}{{{{\left( {1 + x} \right)}^{18}}}}dx \cr
& \cr
& \left( {\text{a}} \right)dy = \frac{{2 - 3x}}{{2\sqrt {1 - x} }}dx{\text{ and }}\left( {\text{b}} \right)\,dy = - \frac{{17}}{{{{\left( {1 + x} \right)}^{18}}}}dx \cr} $$
