Calculus 10th Edition

$\displaystyle \frac{4(\ln x)^{3}}{x}$
By theorem 5.3/1: $\displaystyle \frac{d}{dx}[\ln x]=\frac{1}{x}$ We use the chain rule here: $\left[\begin{array}{ll} u(v)=v^{4} & v(x)=\ln x\\ \frac{du}{dv}=4v^{3} & \frac{dv}{dx}=\frac{1}{x} \end{array}\right]$ $\displaystyle \frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}=4v^{3}\cdot\frac{1}{x}=\frac{4(\ln x)^{3}}{x}$