How do you find the derivative of y=(x+1)/(x-1)y=x+1x1?

2 Answers
Dec 23, 2016

The answer is ==-2/(x-1)^2==2(x1)2

Explanation:

This is a ratio of polynomials.

The derivative is

(u/v)'=(u'v-uv')/(v^2)

Here,

u=x+1, =>, u'=1

v=x-1, =>, v'=1

So,

dy/dx=(1*(x-1)-1(x+1))/(x-1)^2

=(x-1-x-1)/(x-1)^2

=-2/(x-1)^2

Dec 23, 2016

d/(dx) ((x+1)/(x-1)) = frac (-2) ((x-1)^2)

Explanation:

Using the quotient rule:

d/(dx) ((x+1)/(x-1)) = frac ((x-1)*d/(dx) (x+1) - (x+1)*d/(dx)(x-1)) ((x-1)^2)

d/(dx) ((x+1)/(x-1)) = frac ((x-1)*1 - (x+1)*1) ((x-1)^2)

d/(dx) ((x+1)/(x-1)) = frac (x-1 - x-1) ((x-1)^2)

d/(dx) ((x+1)/(x-1)) = frac (-2) ((x-1)^2)