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

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Answer 1 · 2016-12-23T11:02:04

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

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,

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

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

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

Answer 2 · 2016-12-23T11:06:11

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

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

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

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

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

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