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)#