## Calculus: Early Transcendentals (2nd Edition)

$${4^{2x + 1}}{x^{4x}}\left( {\ln 2x + 1} \right)$$
\eqalign{ & f\left( x \right) = {\left( {2x} \right)^{4x}} \cr & {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr & f\left( x \right) = {e^{\ln {{\left( {2x} \right)}^{4x}}}} \cr & f\left( x \right) = {e^{4x\ln \left( {2x} \right)}} \cr & {\text{evaluate the derivative}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {{e^{4x\ln \left( {2x} \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 & f'\left( x \right) = {e^{4x\ln \left( {2x} \right)}}\frac{d}{{dx}}\left( {4x\ln \left( {2x} \right)} \right) \cr & {\text{product rule}} \cr & f'\left( x \right) = {e^{4x\ln \left( {2x} \right)}}\left( {4\ln \left( {2x} \right) + 4x \cdot \frac{1}{x}} \right) \cr & {\text{simplify}} \cr & f'\left( x \right) = {\left( {2x} \right)^{4x}}\left( {4\ln \left( {2x} \right) + 4} \right) \cr & f'\left( x \right) = {2^{4x}}{x^{4x}}\left( {4\ln \left( {2x} \right) + 4} \right) \cr & f'\left( x \right) = {4^{2x}}{x^{4x}}\left( {4\ln \left( {2x} \right) + 4} \right) \cr & f'\left( x \right) = {4^{2x + 1}}{x^{4x}}\left( {\ln 2x + 1} \right) \cr}