Chapter 11 - Section 11.5 - Derivatives of Logarithmic and Exponential Functions - Exercises - Page 842: 23

$ \displaystyle \frac{4x}{2x^{2}+1}$

(see p. 838, Generalized Rule)$\quad$ $\displaystyle \frac{d}{dx}[\ln|u|]=\frac{1}{u}\cdot\frac{du}{dx}$ $u(x)=2x^{2}+1$ $\displaystyle \frac{du}{dx}=2(2x)=4x$ $\displaystyle \frac{d}{dx}[\ln|2x^{2}+1|]=\frac{1}{2x^{2}+1}\cdot(4x)$ $=\displaystyle \frac{4x}{2x^{2}+1}$