## Calculus with Applications (10th Edition)

The correct way to solve the equation is: $D_x(\frac{2x+5}{x^2-1})=\frac{2(x^2-1)-2x(2x+5)}{(x^2-1)^2}$ $=\frac{2x^2-21-4x+10x}{(x^2-1)^2}$ $=\frac{2x^2+6x-21}{(x^2-1)^2}$
We are given $D_x(\frac{2x+5}{x^2-1})=\frac{2x(2x+5)-(x^2-1)2}{(x^2-1)^2}$ $=\frac{4x^2+10x-2x^2+2}{(x^2-1)^2}$ $=\frac{2x^2+10x+2}{(x^2-1)^2}$ Since $D_x(\frac{2x+5}{x^2-1})=\frac{2x(2x+5)-(x^2-1)2}{(x^2-1)^2}$ is incorrect The correct way to solve this equation is: $D_x(\frac{2x+5}{x^2-1})=\frac{2(x^2-1)-2x(2x+5)}{(x^2-1)^2}$ $=\frac{2x^2-21-4x+10x}{(x^2-1)^2}$ $=\frac{2x^2+6x-21}{(x^2-1)^2}$