## Thomas' Calculus 13th Edition

$y'=\frac{-1}{2x^2}(\frac{x^2+x}{x^2})^{-1/2}$
Rewrite the equation: $y=(\frac{x^2+x}{x^2})^{1/2}$ Take the derivative of the equation using Power Rule, Chain Rule, and Quotient Rule: $y'=\frac{1}{2}(\frac{x^2+x}{x^2})^{-1/2}\times\frac{(x^2)(2x+1)-(x^2+x)(2x)}{(x^2)^2}$ $=\frac{1}{2}(\frac{x^2+x}{x^2})^{-1/2}\times\frac{2x^3+x^2-2x^3-2x^2}{x^4}$ $=\frac{1}{2}(\frac{x^2+x}{x^2})^{-1/2}\times\frac{-x^2}{x^4}$ $=\frac{-1}{2x^2}(\frac{x^2+x}{x^2})^{-1/2}$