y=1+x1−xy = \frac{1+x}{1-x}y=1−x1+x
Using quotient rule, vdudx−udvdxv2\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}v2vdxdu−udxdv, we have
dydx=(1−x)(1)−(1+x)(−1)(1−x)2=1−x+1+x(1−x)2\frac{dy}{dx} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^{2}} = \frac{1 - x + 1 + x}{(1-x)^{2}}dxdy=(1−x)2(1−x)(1)−(1+x)(−1)=(1−x)21−x+1+x
= 2(1−x)2\frac{2}{(1-x)^{2}}(1−x)22.
