Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 325: 43

$\displaystyle \frac{2}{x}$

By theorem 5.3/2: $\displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$ Here, u(x)$=x^{2},\displaystyle \ \ \ \frac{du}{dx}=2x$ $\displaystyle \frac{d}{dx}[\ln(x^{2})]=\frac{1}{x^{2}}\cdot 2x=\frac{2}{x}$