# Factoring out quotient of an expression.

I am following the MIT Open Courseware on Single Variable Calculus and in the first lesson when taking the derivative of a simple function I found myself confused because of my knowledge gap in factoring expressions. I'm confused on this particular step.

$$\frac{\delta F}{\delta x} = \frac {\frac{1}{x_{0}+ \delta x} - \frac{1}{x_0}}{ \delta x} = \frac{1}{\delta x}{ (\frac{x_{0} - (x_{0} + \delta x)}{(x_{0} +\delta x){x_{0}}})}$$

How does the factoring out of the quotient happen? Could you explain it step by step?

$\delta_x$ is in the denominator. Therefore it can be factored out. And the common denominator of $x_0$ and $x_0+\delta_x$is the product of both: $x_0\cdot (x_0+\delta_x )$. Therefore $\frac{1}{x_0+\delta_x}$ has to be expanded by $x_0$ and $\frac{1}{x_0}$ has to be expanded by $x_0+\delta_x$:
$\frac{1}{x_0+\delta_x}\cdot \frac{x_0}{x_0}-\frac{1}{x_0} \cdot \frac{x_0+\delta_x}{x_0+\delta_x}=\frac{x_0-(x_0+\delta_x)}{x_0\cdot (x_0+\delta_x )}$
The numerator becomes $-\delta x$. That is what cancels with the $\frac1{\delta x}$ outside the brackets.
Then you are left with $\frac{-1}{(x_0+\delta x)x_0}$, which is continuous at $\delta x=0$, so you can just let $\delta x=0$, and get $\frac{-1}{x_0^2}$