Answer
$$\,\,\,\,\,dy = 0.060$$
Work Step by Step
$$\eqalign{
& y = \sqrt {3x + 2} ;\,\,\,\,\,\,x = 4{\text{ and }}\Delta x = 0.15 \cr
& {\text{The differential }}dy{\text{ is the product of }}f'\left( x \right){\text{ and }}dx,{\text{ or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)dx \cr
& {\text{Find }}f'\left( x \right) \cr
& f'\left( x \right) = \left( {\sqrt {3x + 2} } \right)' \cr
& f'\left( x \right) = \left[ {{{\left( {3x + 2} \right)}^{1/2}}} \right]' \cr
& {\text{by using the chain rule}} \cr
& f'\left( x \right) = \frac{1}{2}{\left( {3x + 2} \right)^{ - 1/2}}\left( 3 \right) \cr
& f'\left( x \right) = \frac{3}{{2\sqrt {3x + 2} }} \cr
& {\text{Then}}{\text{, the differential }}dy{\text{ is}} \cr
& \,\,\,\,\,dy = \frac{3}{{2\sqrt {3x + 2} }}dx \cr
& {\text{As }}dx{\text{ approaches 0}}{\text{, the value of }}dy{\text{ gets closer and closer that of }}\Delta y,{\text{ }} \cr
& {\text{so that for small nonzero values of }}dx, \cr
& \,\,\,\,\,dy \approx \Delta y, \cr
& {\text{or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)\Delta x \cr
& {\text{Then }} \cr
& \,\,\,\,\,dy = \frac{3}{{2\sqrt {3x + 2} }}\Delta x \cr
& {\text{Substituting }}x = 4{\text{ and }}\Delta x = dx = 0.15 \cr
& \,\,\,\,\,dy = \frac{3}{{2\sqrt {3\left( 4 \right) + 2} }}\left( {0.15} \right) \cr
& {\text{simplifying}} \cr
& \,\,\,\,\,dy = 0.060 \cr} $$
