Logarithmic Differentiation of $x(3x^2-9)$

Question

I am trying to teach myself how to use logarithmic differentiation, but I'm not getting the right answer for some reason. I think it is because I may be improperly using log rules, but I can't find anything on how to use them with complicated terms like $3x^2-9$

$$y = x(3x^2-9)$$

Apply $\ln()$ to both sides.

$$\ln(y)=\ln(x(3x^2-9))$$

$$\ln(y)= \ln(x) + 2\ln(3x)-\ln9$$

$$\ln(y)= \ln(x) + 2\ln3 + 2\ln(x)-\ln9$$

Differentiate each term

$$\frac{1}{y}y'=\frac{1}{x}+0+\frac{2}{x}-0$$

Multiply both sides by $y$.

$$y'=(3x^3-9x)\frac{3}{x}$$

But when I use Symbolab to solve the derivative and plug both derivatives into Desmos, I get different graphs.

Answer 1

Here is the correct calculation: if you set $\; f(x)=x(3x^2-9)$, then $$\frac{f'(x)}{f(x)}=\frac 1x+\frac{6x}{3x^2-9}=\frac{3x^2-9+6x^2}{f(x))}=\frac{9(x^2-1)}{f(x)},$$ whence $\;f'(x)=9(x^2-1)$.

Answer 2

The problem is that $$\ln(x\cdot (3x^2-9))=\ln x+\ln(3x^2-9)\color{red}{\not =}\ln x +\ln(3x^2)-\ln 9$$