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

May 19, 2015

Have a look:

May 19, 2015

Firstly, let's remember the limit definition of the derivative :

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

Here, we have the function $f \left(x\right) = \frac{1}{x - 1}$, so :

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

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

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

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

$= - \frac{1}{\left(x + 0 + 1\right) \left(x + 1\right)} = - \frac{1}{x + 1} ^ 2$.

That's it.