## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = - 3{\left( {4x + 3} \right)^4}{\left( {x + 1} \right)^{ - 4}} + 16{\left( {x + 1} \right)^{ - 3}}{\left( {4x + 3} \right)^3}$$
\eqalign{ & y = {\left( {4x + 3} \right)^4}{\left( {x + 1} \right)^{ - 3}} \cr & {\text{differentiate with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {4x + 3} \right)}^4}{{\left( {x + 1} \right)}^{ - 3}}} \right] \cr & {\text{use the product rule}} \cr & \frac{{dy}}{{dx}} = {\left( {4x + 3} \right)^4}\frac{d}{{dx}}\left[ {{{\left( {x + 1} \right)}^{ - 3}}} \right] + {\left( {x + 1} \right)^{ - 3}}\frac{d}{{dx}}\left[ {{{\left( {4x + 3} \right)}^4}} \right] \cr & {\text{use the chain rule }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}\frac{{du}}{{dx}} \cr & \frac{{dy}}{{dx}} = {\left( {4x + 3} \right)^4}\left( { - 3} \right){\left( {x + 1} \right)^{ - 4}}\frac{d}{{dx}}\left[ {x + 1} \right] + {\left( {x + 1} \right)^{ - 3}}\left( 4 \right){\left( {4x + 3} \right)^3}\frac{d}{{dx}}\left[ {4x + 3} \right] \cr & {\text{solve derivatives and simplify}} \cr & \frac{{dy}}{{dx}} = {\left( {4x + 3} \right)^4}\left( { - 3} \right){\left( {x + 1} \right)^{ - 4}}\left( 1 \right) + {\left( {x + 1} \right)^{ - 3}}\left( 4 \right){\left( {4x + 3} \right)^3}\left( 4 \right) \cr & \frac{{dy}}{{dx}} = - 3{\left( {4x + 3} \right)^4}{\left( {x + 1} \right)^{ - 4}} + 16{\left( {x + 1} \right)^{ - 3}}{\left( {4x + 3} \right)^3} \cr}