How do you find the derivative of #f(x)=1/(x-1)# using the limit process?

1 Answer
Steve M
Jan 9, 2017

#f'(x) = - (1 ) /((x-1)^2 #

Explanation:

The definition of the derivative of #y=f(x)# is

# f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #

So if # f(x) = 1/(x-1) # then;

And so the derivative of #f(x)# is given by:

#f'(x) = lim_(h rarr 0) ( 1/((x+h)-1) - 1/(x-1) ) /h #
# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( (x-1)-(x+h-1) ) /( h(x-1)(x+h-1) #
# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( -h ) /( h(x-1)(x+h-1) #
# \ \ \ \ \ \ \ \ = lim_(h rarr 0) - (1 ) /((x-1)(x+h-1) #
# \ \ \ \ \ \ \ \ = - (1 ) /((x-1)(x+0-1) #
# \ \ \ \ \ \ \ \ = - (1 ) /((x-1)^2 #

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