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

Question

Aurora Alvarado · Accepted Answer

There are a few formulas for the definition of a derivative that follow the same idea:

#lim_(h->0)[f(x+h)-f(x)]/h#

#lim_(x->a)[f(x)-f(a)]/(x-a)#

#lim_(Deltax->0)[f(x+Deltax)-f(x)]/(Deltax)#

We'd plug in our function into one of these limits and solve.

Since #f(x)=(x+1)/(x-1)# Then: #lim_(h->0)[[((x+h)+1)/((x+h)-1)]-[(x+1)/(x-1)]]/h = f'(x)# If you expand this algebraically and simplify, you should end up with: #lim_(h->0)-2/((x+h-1)(x-1))# Solve the limit: Now you can plug in #h#: #=-2/((x-1)(x-1))=-2/(x-1)^2=f'(x)#

Elijah Abram · Answer

undefined To find the derivative of $ f(x) = \frac{x+1}{x-1} $ using the definition of a derivative, we can apply the limit definition of the derivative, which states:

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

Substitute $ f(x) = \frac{x+1}{x-1} $ into the definition and simplify:

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

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

$$ = \lim_{h \to 0} \frac{x^2 - x + hx - h - x + 1 - x^2 - hx - x - h + 1}{h(x+h-1)(x-1)} $$

$$ = \lim_{h \to 0} \frac{-2h}{h(x+h-1)(x-1)} $$

$$ = \lim_{h \to 0} \frac{-2}{(x+h-1)(x-1)} $$

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

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