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

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.

$y = {\left(\frac{1 + x}{1 - x}\right)}^{3} = {\left(\left(1 + x\right) {\left(1 - x\right)}^{-} 1\right)}^{3} = {\left(1 + x\right)}^{3} {\left(1 - x\right)}^{-} 3$

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 $\implies y ' = 3 {\left(\frac{1 + x}{1 - x}\right)}^{2} \cdot \left(\frac{1 + x}{1 - x}\right) '$

$y ' = 3 {\left(\frac{1 + x}{1 - x}\right)}^{2} \cdot \frac{\left(1 - x\right) \left(1\right) - \left(1 + x\right) \left(- 1\right)}{1 - x} ^ 2$

$y ' = 3 {\left(\frac{1 + x}{1 - x}\right)}^{2} \cdot \frac{\left(1 - x\right) - \left(1 + x\right) \left(- 1\right)}{1 - x} ^ 2$

$y ' = 3 {\left(\frac{1 + x}{1 - x}\right)}^{2} \cdot \frac{\left(1 - x\right) + \left(1 + x\right)}{1 - x} ^ 2$

$y ' = 3 {\left(\frac{1 + x}{1 - x}\right)}^{2} \cdot \frac{1 - x + 1 + x}{1 - x} ^ 2$

$y ' = 3 {\left(\frac{1 + x}{1 - x}\right)}^{2} \cdot \frac{1 + 1}{1 - x} ^ 2$

$y ' = 3 {\left(\frac{1 + x}{1 - x}\right)}^{2} \cdot \frac{2}{1 - x} ^ 2$

$y ' = 6 {\left(\frac{1 + x}{1 - x}\right)}^{2} \cdot \frac{1}{1 - x} ^ 2$

$y ' = 6 \frac{{\left(1 + x\right)}^{2}}{1 - x} ^ 2 \cdot \frac{1}{1 - x} ^ 2$

$y ' = 6 \frac{{\left(1 + x\right)}^{2}}{1 - x} ^ 4$