## Calculus: Early Transcendentals (2nd Edition)

$${\left( {1 + \frac{4}{x}} \right)^x}\left( {\frac{x}{{x + 4}} + \ln \left( {\frac{{x + 4}}{x}} \right) - 1} \right)$$
\eqalign{ & \frac{d}{{dx}}{\left( {1 + \frac{4}{x}} \right)^x} \cr & {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr & = {e^{\ln {{\left( {1 + \frac{4}{x}} \right)}^x}}} \cr & = {e^{x\ln \left( {1 + \frac{4}{x}} \right)}} \cr & {\text{evaluate the derivative}} \cr & = \frac{d}{{dx}}\left( {{e^{x\ln \left( {1 + \frac{4}{x}} \right)}}} \right) \cr & = \frac{d}{{dx}}\left( {{e^{x\ln \left( {\frac{{x + 4}}{x}} \right)}}} \right) \cr & {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr & = {e^{x\ln \left( {1 + \frac{4}{x}} \right)}}\frac{d}{{dx}}\left( {x\ln \left( {\frac{{x + 4}}{x}} \right)} \right) \cr & = {e^{x\ln \left( {1 + \frac{4}{x}} \right)}}\frac{d}{{dx}}\left( {x\ln \left( {x + 4} \right) - x\ln x} \right) \cr & {\text{product rule}} \cr & = {e^{x\ln \left( {1 + \frac{4}{x}} \right)}}\left( {x\left( {\frac{1}{{x + 4}}} \right) + \ln \left( {x + 4} \right) - x\left( {\frac{1}{x}} \right) - \ln x} \right) \cr & {\text{simplify}} \cr & = {\left( {1 + \frac{4}{x}} \right)^x}\left( {\frac{x}{{x + 4}} + \ln \left( {x + 4} \right) - 1 - \ln x} \right) \cr & = {\left( {1 + \frac{4}{x}} \right)^x}\left( {\frac{x}{{x + 4}} + \ln \left( {\frac{{x + 4}}{x}} \right) - 1} \right) \cr}