Find the derivative of ln(3x^2).

Step-by-step solution

1. 1

   Apply the chain rule for logarithms. For any differentiable inner function,

   $\frac{d}{dx}\ln u=\frac{u'}{u}$

   Here $u=3x^2$, so the derivative is the derivative of the inside divided by the inside — a single formula, no product rule needed.

2. 2

   Differentiate the inside.

   $u=3x^2\ \Longrightarrow\ u'=6x$

3. 3

   Form the quotient and simplify.

   $\frac{d}{dx}\ln\left(3x^2\right)=\frac{6x}{3x^2}=\frac{2}{x}$

   Both the coefficient $3$ and one power of $x$ cancel. The answer is valid for $x\neq 0$, which is also where $\ln(3x^2)$ is defined.

4. 4

   See why the 3 had to vanish. Log rules give

   $\ln\left(3x^2\right)=\ln 3+2\ln x$

   a constant plus $2\ln x$. Constants differentiate to zero, so the derivative is $2\cdot\tfrac1x=\tfrac2x$ — matching, and explaining, the cancellation. Any multiplicative constant inside a logarithm is invisible to the derivative.

5. 5

   Check numerically. At $x=2.3$ a central difference gives $0.8695652$, and $\tfrac{2}{2.3}=0.8695652$ ✓. Note the derivative is positive for $x>0$ and negative for $x<0$, matching the U-shaped domain of $\ln(3x^2)$.