Calculus (3rd Edition)

$$f'(x)=\frac{8x}{4x^2+1}.$$
Recall that $(\ln x)'=\dfrac{1}{x}$ Since $f(x)= \ln(4x^2+1)$, then the derivative is given by $$f'(x)=\frac{1}{4x^2+1} (4x^2+1)'=\frac{8x}{4x^2+1}.$$