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

Oct 3, 2014

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

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

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

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

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

Find the least common denominator

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

Distribute the negative in the numerator of the complex fraction

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

Simplify the numerator of the complex fraction

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

Division is equivalent to multiplying by the reciprocal

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

Cross cancel the $h$ factors

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

Substitute in the value of 0 for $h$ and simplify

$= \frac{1}{\left(1 - x - 0\right) \left(1 - x\right)}$

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

$= \frac{1}{1 - x} ^ 2$