How do you find the derivative of #y= ((1+x)/(1-x))^3# ?

1 Answer
Sep 25, 2014

We can do a number of things ..

1) Use the chain rule and quotient rule

2) Use the chain rule and the power rule after the following transformations.

#y=((1+x)/(1-x))^3=((1+x)(1-x)^-1)^3=(1+x)^3(1-x)^-3#

3) You could multiply out everything, which takes a bunch of time, and then just use the quotient rule.

Let's keep it simple and just use the chain rule and quotient rule.

Chain Rule #=>y'=3((1+x)/(1-x))^2*((1+x)/(1-x))'#

#y'=3((1+x)/(1-x))^2*((1-x)(1)-(1+x)(-1))/(1-x)^2#

#y'=3((1+x)/(1-x))^2*((1-x)-(1+x)(-1))/(1-x)^2#

#y'=3((1+x)/(1-x))^2*((1-x)+(1+x))/(1-x)^2#

#y'=3((1+x)/(1-x))^2*(1-x+1+x)/(1-x)^2#

#y'=3((1+x)/(1-x))^2*(1+1)/(1-x)^2#

#y'=3((1+x)/(1-x))^2*(2)/(1-x)^2#

#y'=6((1+x)/(1-x))^2*(1)/(1-x)^2#

#y'=6((1+x)^2)/(1-x)^2*(1)/(1-x)^2#

#y'=6((1+x)^2)/(1-x)^4#