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

2 Answers
Apr 26, 2016

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

Explanation:

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#

Apr 26, 2016

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

Explanation:

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#