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#