Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 325: 59

$-\frac{4}{x(4+x^2)}$

$f(x)=ln(\frac{\sqrt{4+x^2}}{x})$ $=ln(4+x^2)^\frac{1}{2}-ln x$ $=\frac{1}{2}ln(4+x^2)^\frac{1}{2}-ln x$ $f'(x)=(\frac{!}{2})(\frac{2x}{4+x^2})-(\frac{1}{x^2})$ $=\frac{x}{4+x^2}-\frac{1}{x}$ $=\frac{x^2-4-x^2}{x(4+x^2)}$ $=-\frac{4}{x(4+x^2)}$