How do you use the formal definition of differentiation as a limit to find the derivative of #f(x)=1/(x-1)#?

Question

Christopher Allers · Accepted Answer

Have a look:

Luke Alvarez · Answer

Firstly, let's remember the limit definition of the derivative :

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

Here, we have the function #f(x) = 1/(x-1)#, so :

#f'(x)=lim_(h->0)(1/(x+h+1) - 1/(x+1))/h =lim_(h->0)(((x+1)-(x+h+1))/((x+h+1)(x+1)))/h#

#= lim_(h->0)((x+1)-(x+h+1))/(h(x+h+1)(x+1))#

#= lim_(h->0)((-h))/(h(x+h+1)(x+1))#

#= lim_(h->0)-1/((x+h+1)(x+1))#

#= -1/((x+0+1)(x+1)) = -1/(x+1)^2#.

That's it.

Isaiah Allen · Answer

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

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

2. Substitute $ f(x) = \frac{1}{x-1} $ into the definition.

3. Simplify the expression using algebraic manipulation.

4. Take the limit as $ h $ approaches 0 to find the derivative.

Applying these steps will yield the derivative of $ f(x) = \frac{1}{x-1} $ using the formal definition of differentiation.

Axel Alvarez · Answer

undefined To use the formal definition of differentiation as a limit to find the derivative of $ f(x) = \frac{1}{x - 1} $, we start with the definition of the derivative:

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

Substitute $ f(x) = \frac{1}{x - 1} $ into this definition:

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

Combine the fractions:

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

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

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

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

Now, we can evaluate the limit as $ h $ approaches 0:

$$ f'(x) = \frac{-1}{(x - 1)^2} $$

Therefore, the derivative of $ f(x) = \frac{1}{x - 1} $ with respect to $ x $ is $ f'(x) = \frac{-1}{(x - 1)^2} $.