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

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Answer 1 · 2014-10-03T03:44:35

$f'(x)=lim_(h->0) (f(x-h)-f(x))/h$

$f(x)=1/(1-x)$

$f(x+h)=1/(1-(x+h))=1/(1-x-h)$

$f'(x)=lim_(h->0) (1/(1-x-h)-1/(1-x))/h$

$f'(x)=lim_(h->0) (1/(1-x-h) * (1-x)/(1-x) -1/(1-x)*(1-x-h)/(1-x-h))/h$

Find the least common denominator

$f'(x)=lim_(h->0) ((1-x)/((1-x-h)(1-x))-(1-x-h)/((1-x-h)(1-x)))/h$

Distribute the negative in the numerator of the complex fraction

$f'(x)=lim_(h->0) ((1-x-1+x+h)/((1-x-h)(1-x)))/h$

Simplify the numerator of the complex fraction

$f'(x)=lim_(h->0) ((h)/((1-x-h)(1-x)))/h$

Division is equivalent to multiplying by the reciprocal

$f'(x)=lim_(h->0) (h)/((1-x-h)(1-x))*1/h$

Cross cancel the $h$ factors

$f'(x)=lim_(h->0) (1)/((1-x-h)(1-x))$

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

$=(1)/((1-x-0)(1-x))$

$=(1)/((1-x)(1-x))$

$=1/(1-x)^2$

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