I need to take the derivative of $\displaystyle Y=(1+ln7x)^3$
I get $\displaystyle 3(1+ln7x)^2 ({\color{red}1+\frac{7}{x}})$
Is this correct?
Almost, but not quite, because the derivative of $\displaystyle 1+\ln(7x)$ happens to be $\displaystyle 0+\frac{1}{7x}\cdot 7=\frac{1}{x}$, and not what you thought it is.