## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{d\theta }} = \frac{{\pi {{\left( {\ln \theta } \right)}^{\pi - 1}}}}{\theta }$$
\eqalign{ & y = {\left( {\ln \theta } \right)^\pi } \cr & {\text{find the derivative of }}y{\text{ with respect to }}\theta \cr & \frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{{\left( {\ln \theta } \right)}^\pi }} \right] \cr & {\text{Use the general power rule for differentiation }}\cr & \frac{d}{{d\theta }}\left[ {{u^n}} \right] = n{u^{n - 1}}\frac{{du}}{{d\theta }}{\text{ }} \cr & {\text{for this exercise let }}n = \pi {\text{ and }}u = \ln \theta {\text{; then}}{\text{,}} \cr & \frac{{dy}}{{d\theta }} = \pi {\left( {\ln \theta } \right)^{\pi - 1}}\frac{d}{{d\theta }}\left[ {\ln \theta } \right] \cr & {\text{solve the derivative}} \cr & \frac{{dy}}{{d\theta }} = \pi {\left( {\ln \theta } \right)^{\pi - 1}}\left( {\frac{1}{\theta }} \right) \cr & {\text{simplify}} \cr & \frac{{dy}}{{d\theta }} = \frac{{\pi {{\left( {\ln \theta } \right)}^{\pi - 1}}}}{\theta } \cr}