# Difference quotient!

• Aug 4th 2009, 12:05 PM
rowe
Difference quotient!
Need to show that $\frac{\frac{1}{x}-\frac{1}{a}}{x-a} = -\frac{1}{ax}$.

Basically, I'll do something like $\left(\frac{1}{x}-\frac{1}{a}\right)\frac{1}{x-a}$, and manipulate it in several different ways, all of which are far more complicated than the simplified form, like $\frac{a-x}{ax^2-a^2x}$. What's the trick here?
• Aug 4th 2009, 12:32 PM
e^(i*pi)
Quote:

Originally Posted by rowe
Need to show that $\frac{\frac{1}{x}-\frac{1}{a}}{x-a} = -\frac{1}{ax}$.

Basically, I'll do something like $\left(\frac{1}{x}-\frac{1}{a}\right)\frac{1}{x-a}$, and manipulate it in several different ways, all of which are far more complicated than the simplified form, like $\frac{a-x}{ax^2-a^2x}$. What's the trick here?

Note that $ax^2-a^2x = ax(x-a) and (x-a) = -(a-x)$
• Aug 5th 2009, 03:47 AM
HallsofIvy
Quote:

Originally Posted by rowe
Need to show that $\frac{\frac{1}{x}-\frac{1}{a}}{x-a} = -\frac{1}{ax}$.

Basically, I'll do something like $\left(\frac{1}{x}-\frac{1}{a}\right)\frac{1}{x-a}$, and manipulate it in several different ways, all of which are far more complicated than the simplified form, like $\frac{a-x}{ax^2-a^2x}$. What's the trick here?

Actually, the first thing I would have done is get rid of the "compound" fraction by multiplying numerator and denominator on the left by ax:
$\frac{ax(\frac{1}{x}- \frac{1}{a})}{ax(x- a)}$ which immediately gives your $\frac{a-x}{ax(x-a)}$ and then cancel:
$\frac{-(x-a)}{ax(x-a)}= -\frac{1}{ax}$