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

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Answer 1 · 2014-09-25T23:33:08

We can do a number of things ..

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

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

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

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

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

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

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

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

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

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

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