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

calculusderivativeslogarithms

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.

Best Answer

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

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