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

Question

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

Emilia Aguilar · Accepted Answer

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

We can use the Quotient Rule to find the derivative, which states:

#y'=(g(x)h'(x)-h(x)g'(x))/(g(x))^2#

Where in our case,

#h(x)=x# and #g(x) = x-1#

We then need to find the derivative of each of these:

#h'(x) = 1# and #g'(x)=1#

Reassembling this we find our derivative:

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

Ezekiel Alexander · Answer

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

In a previous answer, we used the quotient rule which is a simplification of the product rule. Some people, including me, don't like remembering more than one rule, so let's do that again using the product rule , which states:

#f'(x) = g'(x)h(x) + g(x)h'(x)#

in our case

#g(x) = x# and #h(x) = (x-1)^(-1)#

Now we need to find the derivatives of these functions:

#g'(x) = 1# and #h'(x) = -(x-1)^(-2)#

So reassembling we find our solution:

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

Emilia Aguilar · Answer

undefined To find the derivative of $ f(x) = \frac{x}{x-1} $, you can use the quotient rule, which states that if $ f(x) = \frac{g(x)}{h(x)} $, then $ f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2} $. Applying this rule to $ f(x) $, where $ g(x) = x $ and $ h(x) = x - 1 $, we get:

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