## Calculus 10th Edition

$f'(x)= \frac{x\cos(x)-\sin(x)}{x^{2}}$ $g'(x)= \frac{x\cos(x)-\sin(x)}{x^{2}}$ The slope of $f(x)$ is equal to the slope of $g(x)$. They are parallel functions.
Quotient rule: $\frac{d}{dx}$$\frac{f(x)}{g(x)}$=$\frac{g(x)f'(x)-f(x)g'(x)}{g(x)g(x)}$ $f(x)= \frac{sin(x)-3x}{x}$ $f'(x)= \frac{(x)[cos(x)-3]-[sin(x)-3x](1)}{x^{2}}$=$\frac{xcos(x)-3x-sin(x)+3x}{x^{2}}$=$\frac{xcos(x)-sin(x)}{x^{2}}$ $g(x)=\frac{sin(x)+2x}{x}$ $g'(x)=\frac{(x)[cos(x)+2]-[sin(x)-2x](1)}{x^{2}}$=$\frac{cos(x)+2x-sin(x)-2x}{x^{2}}$=$\frac{xcos(x)-sin(x)}{x^{2}}$ The slope of $f(x)$ is equal to the slope of $g(x)$. They are parallel functions.