## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = {\left( {x + 2} \right)^{x + 2}}\left( {1 + \ln \left( {x + 2} \right)} \right)$$
\eqalign{ & y = {\left( {x + 2} \right)^{x + 2}} \cr & {\text{take the natural logarithm }} \cr & \ln y = \ln {\left( {x + 2} \right)^{x + 2}} \cr & {\text{use the power property for logarithms}} \cr & \ln y = \left( {x + 2} \right)\ln \left( {x + 2} \right) \cr & {\text{differentiate both sides with respecto to }}x \cr & \frac{d}{{dx}}\left( {\ln y} \right) = \frac{d}{{dx}}\left[ {\left( {x + 2} \right)\ln \left( {x + 2} \right)} \right] \cr & {\text{use the product rule}} \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \left( {x + 2} \right)\frac{d}{{dx}}\left[ {\ln \left( {x + 2} \right)} \right] + \ln \left( {x + 2} \right)\frac{d}{{dx}}\left[ {\left( {x + 2} \right)} \right] \cr & \frac{1}{y}\frac{{dy}}{{dx}} = \left( {x + 2} \right)\left( {\frac{1}{{x + 2}}} \right) + \ln \left( {x + 2} \right)\left( 1 \right) \cr & \frac{1}{y}\frac{{dy}}{{dx}} = 1 + \ln \left( {x + 2} \right) \cr & {\text{solve for }}\frac{{dy}}{{dx}} \cr & \frac{{dy}}{{dx}} = y\left( {1 + \ln \left( {x + 2} \right)} \right) \cr & {\text{replace }}y = {\left( {x + 2} \right)^{x + 2}} \cr & \frac{{dy}}{{dx}} = {\left( {x + 2} \right)^{x + 2}}\left( {1 + \ln \left( {x + 2} \right)} \right) \cr}