# Calculus Derivatives

• January 28th 2009, 04:50 PM
SoYah
Calculus Derivatives
Find dy/dx, where y = ((x+2)(x+4))/(x-1)
• January 28th 2009, 05:56 PM
chabmgph
Quote:

Originally Posted by SoYah
Find dy/dx, where y = ((x+2)(x+4))/(x-1)

The key here is the quotient rule:
$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}$

Since $y = \frac{(x+2)(x+4)}{(x-1)}=\frac{x^2+6x+8}{(x-1)}$,

we have $\frac{dy}{dx}=\frac{[(x^2+6x+8)'](x-1)-(x^2+6x+8)[(x-1)']}{(x-1)^2}$ $=\frac{(2x+6)(x-1)-(x^2+6x+8)(1)}{(x-1)^2}=...$

I believe you can take it from here.