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

Jun 11, 2018

$\frac{d}{\mathrm{dx}} \left(\frac{x + 1}{x - 1}\right) = - \frac{2}{x - 1} ^ 2$

#### Explanation:

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

so:

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

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

$\frac{\mathrm{df}}{\mathrm{dx}} = {\lim}_{h \to 0} \frac{1}{h} \frac{\textcolor{b l u e}{\left(x + 1\right) \left(x - 1\right)} + h \left(x - 1\right) - \textcolor{b l u e}{\left(x - 1\right) \left(x + 1\right)} - h \left(x + 1\right)}{\left(x + h - 1\right) \left(x - 1\right)}$

$\frac{\mathrm{df}}{\mathrm{dx}} = {\lim}_{h \to 0} \frac{1}{h} \frac{\textcolor{b l u e}{h x} - h - \textcolor{b l u e}{h x} - h}{\left(x + h - 1\right) \left(x - 1\right)}$

$\frac{\mathrm{df}}{\mathrm{dx}} = {\lim}_{h \to 0} \frac{1}{\textcolor{b l u e}{h}} \frac{- 2 \textcolor{b l u e}{h}}{\left(x + h - 1\right) \left(x - 1\right)}$

$\frac{\mathrm{df}}{\mathrm{dx}} = {\lim}_{h \to 0} \frac{- 2}{\left(x + h - 1\right) \left(x - 1\right)} = - \frac{2}{x - 1} ^ 2$