## Thomas' Calculus 13th Edition

$$4x{\left( {{x^2}\ln x} \right)^3}\left( {1 + 2\ln x} \right)$$
\eqalign{ & y = {\left( {{x^2}\ln x} \right)^4} \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {{x^2}\ln x} \right)}^4}} \right] \cr & {\text{use the chain rule}} \cr & \frac{{dy}}{{dx}} = 4{\left( {{x^2}\ln x} \right)^{4 - 1}}\frac{d}{{dx}}\left[ {{x^2}\ln x} \right] \cr & \frac{{dy}}{{dx}} = 4{\left( {{x^2}\ln x} \right)^3}\frac{d}{{dx}}\left[ {{x^2}\ln x} \right] \cr & {\text{use the product rule}} \cr & \frac{{dy}}{{dx}} = 4{\left( {{x^2}\ln x} \right)^3}\left( {{x^2}\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ {{x^2}} \right]} \right) \cr & {\text{solve the derivatives }} \cr & \frac{{dy}}{{dx}} = 4{\left( {{x^2}\ln x} \right)^3}\left( {{x^2}\left( {\frac{1}{x}} \right) + \ln x\left( {2x} \right)} \right) \cr & {\text{simplifying, we get:}} \cr & \frac{{dy}}{{dx}} = 4{\left( {{x^2}\ln x} \right)^3}\left( {x + 2x\ln x} \right) \cr & \frac{{dy}}{{dx}} = 4x{\left( {{x^2}\ln x} \right)^3}\left( {1 + 2\ln x} \right) \cr}