## Calculus with Applications (10th Edition)

$$\,\,\,\,\,dy = 0.036706$$
\eqalign{ & y = \sqrt {4x - 1} ;\,\,\,\,\,\,x = 5{\text{ and }}\Delta x = 0.08 \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 {4x - 1} } \right)' \cr & f'\left( x \right) = \left[ {{{\left( {4x - 1} \right)}^{1/2}}} \right]' \cr & {\text{by using the chain rule}} \cr & f'\left( x \right) = \frac{1}{2}{\left( {4x - 1} \right)^{ - 1/2}}\left( 4 \right) \cr & f'\left( x \right) = \frac{2}{{\sqrt {4x - 1} }} \cr & {\text{Then}}{\text{, the differential }}dy{\text{ is}} \cr & \,\,\,\,\,dy = \frac{2}{{\sqrt {4x - 1} }}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{2}{{\sqrt {4x - 1} }}\Delta x \cr & {\text{Substituting }}x = 5{\text{ and }}\Delta x = dx = 0.08 \cr & \,\,\,\,\,dy = \frac{2}{{\sqrt {4\left( 5 \right) - 1} }}\left( {0.08} \right) \cr & {\text{simplifying}} \cr & \,\,\,\,\,dy = 0.036706 \cr}