# How do you use the definition of a derivative to find the derivative of f(x)=(x+1)/(x-1)?

Dec 11, 2016

There are a few formulas for the definition of a derivative that follow the same idea:

${\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

${\lim}_{x \to a} \frac{f \left(x\right) - f \left(a\right)}{x - a}$

${\lim}_{\Delta x \to 0} \frac{f \left(x + \Delta x\right) - f \left(x\right)}{\Delta x}$

We'd plug in our function into one of these limits and solve.

#### Explanation:

Since $f \left(x\right) = \frac{x + 1}{x - 1}$

Then:

${\lim}_{h \to 0} \frac{\left[\frac{\left(x + h\right) + 1}{\left(x + h\right) - 1}\right] - \left[\frac{x + 1}{x - 1}\right]}{h} = f ' \left(x\right)$

If you expand this algebraically and simplify, you should end up with:

${\lim}_{h \to 0} - \frac{2}{\left(x + h - 1\right) \left(x - 1\right)}$

Solve the limit:

Now you can plug in $h$:

$= - \frac{2}{\left(x - 1\right) \left(x - 1\right)} = - \frac{2}{x - 1} ^ 2 = f ' \left(x\right)$