Find y' for y = ln x²

Find $y'$ for $y = \ln x^{2}$.

Step 1: Simplify using the logarithm laws

The function is defined for every $x \neq 0$, since $x^{2} \gt 0$ there. Because $\ln x^{2} = 2\ln|x|$ (the absolute value is needed to cover negative $x$): $y = 2\ln|x|, \qquad x \neq 0$.

Step 2: Differentiate

Using $\bigl(\ln|x|\bigr)' = \dfrac{1}{x}$, valid on both sides of the origin: $y' = 2\cdot\frac{1}{x} = \frac{2}{x}$.

Step 3: Verification

Differentiating directly as a composition, with $u = x^{2}$ and $u' = 2x$: \[y' = \frac{1}{u}\cdot u' = \frac{1}{x^{2}}\cdot 2x = \frac{2}{x} \;\;\checkmark\]

\[x = -1:\quad y' = -2; \qquad \frac{y(-1+10^{-5}) - y(-1-10^{-5})}{2\times 10^{-5}} \approx -2.0 \;\;\checkmark\]