Answer to Question #294482 in Calculus for Ianidy

Question

find dy/dx when y= $(1+x/1-x)$ ^-2

Answer 1

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

Let $\displaystyle p=\frac{1+x}{1-x},$ then $\displaystyle y=p^2$

From $\displaystyle y=p^2,\ \frac{dy}{dp}=2p$.

Also, from $\displaystyle p=\frac{1+x}{1-x},\ \frac{dp}{dx}=\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^2}=\frac{2}{(1-x)^2}$, by quotient rule.

Thus by chain rule, we have that;

$\displaystyle \frac{dy}{dx}=\frac{dy}{dp}\cdot\frac{dp}{dx}=2p\cdot\frac{2}{(1-x)^2}=2\left(\frac{1+x}{1-x}\right)\cdot\frac{2}{(1-x)^2}=\frac{4(1+x)}{(1-x)^3}$