How do you use the limit definition to find the derivative of $y = - \frac{1}{x - 1}$ $y=-1/(x-1)$ ?

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Answer 1 · 2016-08-03T11:01:12

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

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

here

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

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

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

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

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

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

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

How do you use the limit definition to find the derivative of y=-1/(x-1)y=−1x−1y=-1/(x-1)?