Answer
$$y'=4x(x^2\ln x)^3(2\ln x+1)$$
Work Step by Step
$$y=(x^2\ln x)^4$$
Recall the following Derivative Rules: $$\frac{d}{dx}(\ln x)=\frac{1}{x}$$
Therefore, we have $$y'=\Big((x^2\ln x)^4\Big)'=4(x^2\ln x)^3(x^2\ln x)'$$
$$y'=4(x^2\ln x)^3\Big((x^2)'\ln x+x^2(\ln x)'\Big)$$
$$y'=4(x^2\ln x)^3\Big(2x\ln x+x^2\times\frac{1}{x}\Big)$$
$$y'=4(x^2\ln x)^3(2x\ln x+x)$$
$$y'=4x(x^2\ln x)^3(2\ln x+1)$$
