## Thomas' Calculus 13th Edition

$${x^3}\ln x$$
\eqalign{ & y = \frac{{{x^4}}}{4}\ln x - \frac{{{x^4}}}{{16}} \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{4}\ln x - \frac{{{x^4}}}{{16}}} \right] \cr & {\text{sum rule for derivatives}} \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{4}\ln x} \right] - \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{{16}}} \right] \cr & {\text{use the product rule for }}\frac{d}{{dx}}\left[ {\frac{{{x^4}}}{4}\ln x} \right] \cr & \frac{{dy}}{{dx}} = \frac{{{x^4}}}{4}\frac{d}{{dx}}\left[ {\ln x} \right] + \ln x\frac{d}{{dx}}\left[ {\frac{{{x^4}}}{4}} \right] - \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{{16}}} \right] \cr & {\text{solve the derivatives and simplify}} \cr & \frac{{dy}}{{dx}} = \frac{{{x^4}}}{4}\left( {\frac{1}{x}} \right) + \ln x\left( {\frac{{4{x^3}}}{4}} \right) - \frac{{4{x^3}}}{{16}} \cr & {\text{simplifying, we get:}} \cr & \frac{{dy}}{{dx}} = \frac{{{x^3}}}{4} + {x^3}\ln x - \frac{{{x^3}}}{4} \cr & \frac{{dy}}{{dx}} = {x^3}\ln x \cr}