Derive $\frac{1}{x}$

Question

I try to derive $\frac{1}{x}$.

I try to do it with the $x_0$ method.

I have $m=\frac {f(x_0)-f(x)}{x_0-x}$ I am using $\frac {1}{x} $ to replace $f(x_0)$ and $f(x)$.

Now we have $m=\frac {\frac {1}{x_0}-\frac {1}{x}}{x_0-x} $

However I want to derive so I need to shorten $x_0-x$ in the denominator. So I need to have anothe $x_0-x$ in the numerator.

I basically need to do: $\left(\frac {1}{x_0}-\frac {1}{x}\right)÷(x_0-x)$

However here I get stuck. It would be nice if someone could help me to calculate this division.

Answer 1

$$\lim_{x \to x_0} \frac{\frac{1}{x}- \frac{1}{x_0}}{x - x_0} = \lim_{x \to x_0} \frac{\frac{x_0 - x}{xx_0}}{x-x_0} = -\lim_{x \to x_0} \frac{x-x_0}{xx_0(x-x_0)} = -\lim_{x \to x_0} \frac{1}{xx_0} = \frac{-1}{x_0^2}$$

Answer 2

Add the fractions:

$$\frac{1}{x_0}-\frac{1}{x} = \frac{x-x_0}{xx_0} = \frac{-(x_0-x)}{xx_0}.$$

Now you can easily cancel the factor you want.

Answer 3

Make a common denominator in the numerator,

$$\frac{\frac{x}{x_0x}-\frac{x_0}{x_0x}}{x_0-x}$$

Now simplify.

Answer 4

$$\frac{1}{x_0}-\frac{1}{x}=\frac{x-x_0}{xx_0}$$