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

Question

Scarlett Allison · Accepted Answer

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

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 #

Adam Adkins · Answer

undefined To find the derivative of $ f(x) = \frac{1}{x - 1} $ using the limit process, follow these steps:

1. Begin with the definition of the derivative:
   $ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $

2. Substitute the function $ f(x) = \frac{1}{x - 1} $ into the definition:
   $ f'(x) = \lim_{h \to 0} \frac{\frac{1}{x + h - 1} - \frac{1}{x - 1}}{h} $

3. Combine the fractions into a single fraction:
   $ f'(x) = \lim_{h \to 0} \frac{(x - 1) - (x + h - 1)}{h(x - 1)(x + h - 1)} $

4. Simplify the numerator:
   $ f'(x) = \lim_{h \to 0} \frac{x - 1 - x - h + 1}{h(x - 1)(x + h - 1)} $
   $ f'(x) = \lim_{h \to 0} \frac{-h}{h(x - 1)(x + h - 1)} $

5. Cancel out the common factor of $ h $:
   $ f'(x) = \lim_{h \to 0} \frac{-1}{(x - 1)(x + h - 1)} $

6. Evaluate the limit as $ h $ approaches 0:
   $ f'(x) = \frac{-1}{(x - 1)(x - 1)} $

7. Simplify the expression:
   $ f'(x) = \frac{-1}{(x - 1)^2} $

Therefore, the derivative of $ f(x) = \frac{1}{x - 1} $ using the limit process is $ f'(x) = \frac{-1}{(x - 1)^2} $.