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

Question

Emilia Aguilar · Accepted Answer

#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 #

Emma Aldridge · Answer

undefined To find the derivative of $ y = -\frac{1}{x - 1} $ using the limit definition, follow these steps:

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

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

3. Simplify the expression inside the limit:
$$ f'(x) = \lim_{h \to 0} \frac{-\frac{1}{x + h - 1} + \frac{1}{x - 1}}{h} $$

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

5. Find a common denominator and combine the fractions:
$$ f'(x) = \lim_{h \to 0} \frac{(x + h - 1) - (x - 1)}{h(x + h - 1)(x - 1)} $$

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

$$ f'(x) = \lim_{h \to 0} \frac{h}{h(x + h - 1)(x - 1)} $$

7. Cancel out $h$ in the numerator and denominator:
$$ f'(x) = \lim_{h \to 0} \frac{1}{(x + h - 1)(x - 1)} $$

8. Evaluate the limit as $h$ approaches 0:
$$ f'(x) = \frac{1}{(x - 1)^2} $$

So, the derivative of $ y = -\frac{1}{x - 1} $ is $ f'(x) = \frac{1}{(x - 1)^2} $.