Question #294482

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


1
Expert's answer
2022-02-07T14:21:10-0500

y=(1+x1x)2\displaystyle y=\left(\frac{1+x}{1-x}\right)^2

Let p=1+x1x,\displaystyle p=\frac{1+x}{1-x}, then y=p2\displaystyle y=p^2

From y=p2, dydp=2p\displaystyle y=p^2,\ \frac{dy}{dp}=2p.

Also, from p=1+x1x, dpdx=(1x)(1)(1+x)(1)(1x)2=2(1x)2\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;

dydx=dydpdpdx=2p2(1x)2=2(1+x1x)2(1x)2=4(1+x)(1x)3\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}


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