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

Question

Elijah Abram · Accepted Answer

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#

Christopher Agresta · Answer

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

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)#

Charles Anctil · Answer

undefined To find the derivative of $ y = \frac{x+1}{x-1} $, you can use the quotient rule, which states that if you have a function $ y = \frac{u}{v} $, then its derivative is given by $ y' = \frac{u'v - uv'}{v^2} $. Applying this rule to the given function, where $ u = x + 1 $ and $ v = x - 1 $, we have $ u' = 1 $ and $ v' = 1 $. Substituting these values into the quotient rule formula, we get:

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

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

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