Answer
$$y=x^3\ln x$$
Work Step by Step
$$y=\frac{x^4}{4}\ln x-\frac{x^4}{16}$$
Recall the following Derivative Rules: $$\frac{d}{dx}(\ln u)=\frac{1}{u}\frac{du}{dx}$$
Therefore, we have $$y'=\Big(\frac{x^4}{4}\ln x\Big)'-\Big(\frac{x^4}{16}\Big)'=\Big(\frac{x^4}{4}\Big)'\ln x+\Big(\frac{x^4}{4}\Big)(\ln x)'-\frac{4x^3}{16}$$
$$y'=\frac{4x^3}{4}\ln x+\frac{x^4}{4}\times\frac{1}{x}-\frac{x^3}{4}$$
$$y'=x^3\ln x+\frac{x^3}{4}-\frac{x^3}{4}=x^3\ln x$$
