find dy/dx when y= (1+x/1−x)(1+x/1-x)(1+x/1−x) -2
y=(1+x1−x)2\displaystyle y=\left(\frac{1+x}{1-x}\right)^2y=(1−x1+x)2
Let p=1+x1−x,\displaystyle p=\frac{1+x}{1-x},p=1−x1+x, then y=p2\displaystyle y=p^2y=p2
From y=p2, dydp=2p\displaystyle y=p^2,\ \frac{dy}{dp}=2py=p2, dpdy=2p.
Also, from p=1+x1−x, dpdx=(1−x)(1)−(1+x)(−1)(1−x)2=2(1−x)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}p=1−x1+x, dxdp=(1−x)2(1−x)(1)−(1+x)(−1)=(1−x)22, by quotient rule.
Thus by chain rule, we have that;
dydx=dydp⋅dpdx=2p⋅2(1−x)2=2(1+x1−x)⋅2(1−x)2=4(1+x)(1−x)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}dxdy=dpdy⋅dxdp=2p⋅(1−x)22=2(1−x1+x)⋅(1−x)22=(1−x)34(1+x)
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