## Calculus with Applications (10th Edition)

The correct way to solve this equation is: $D_x(\frac{x^{2}-4}{x^3})=\frac{2x(x^3)-3x^2(x^2-4)}{(x^3)^2}$ $=\frac{2x^4-3x^4-12x^2}{x^6}$ $=\frac{-x^4-12x^2}{x^6}$ $=\frac{-x^2(x^2+12)}{x^6}$ $=\frac{-x^2-12}{x^4}$
The given equation apply the wrong formula with $D_x(\frac{x}{y})=\frac{x'.y-x.y'}{y^2}$ The correct way to solve this equation is: $D_x(\frac{x^{2}-4}{x^3})=\frac{2x(x^3)-3x^2(x^2-4)}{(x^3)^2}$ $=\frac{2x^4-3x^4-12x^2}{x^6}$ $=\frac{-x^4-12x^2}{x^6}$ $=\frac{-x^2(x^2+12)}{x^6}$ $=\frac{-x^2-12}{x^4}$