1. ## Difference quotient!

Need to show that $\displaystyle \frac{\frac{1}{x}-\frac{1}{a}}{x-a} = -\frac{1}{ax}$.

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

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

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

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

Basically, I'll do something like $\displaystyle \left(\frac{1}{x}-\frac{1}{a}\right)\frac{1}{x-a}$, and manipulate it in several different ways, all of which are far more complicated than the simplified form, like $\displaystyle \frac{a-x}{ax^2-a^2x}$. 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:
$\displaystyle \frac{ax(\frac{1}{x}- \frac{1}{a})}{ax(x- a)}$ which immediately gives your $\displaystyle \frac{a-x}{ax(x-a)}$ and then cancel:
$\displaystyle \frac{-(x-a)}{ax(x-a)}= -\frac{1}{ax}$