A) \[\frac{2}{{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}}\]
B) \[\frac{1}{{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}}\]
C) \[\frac{1}{2{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}}\]
D) \[\frac{1}{{{(1+x)}^{3/2}}{{(1-x)}^{1/2}}}\]
Correct Answer: B
Solution :
\[y=\sqrt{\frac{1+x}{1-x}}\]Þ\[y=\sqrt{\frac{(1+x)(1+x)}{(1-x)(1+x)}}=\sqrt{\frac{{{(1+x)}^{2}}}{1-{{x}^{2}}}}\] Differentiating with respect to x, we get \[\frac{dy}{dx}=\frac{{{(1-x)}^{1/2}}\frac{d}{dx}{{(1+x)}^{1/2}}-{{(1+x)}^{1/2}}\frac{d}{dx}{{(1-x)}^{1/2}}}{(1-x)}\] \[=\frac{(1-x)+(1+x)}{2{{(1-x)}^{3/2}}{{(1+x)}^{1/2}}}\] \[x=\frac{2}{3}y\]. Trick : \[\log y=\frac{1}{2}\log (1+x)-\frac{1}{2}\log (1-x)\] Þ \[\frac{1}{y}\frac{dy}{dx}=\frac{1}{2(1+x)}+\frac{1}{2(1-x)}\] Þ \[\frac{dy}{dx}=\frac{1}{(1+x)(1-x)}\times \sqrt{\frac{1+x}{1-x}}\]\[=\frac{1}{{{(1+x)}^{1/2}}{{(1-x)}^{3/2}}}\].
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