A) \[\frac{4}{{{(x+1)}^{2}}}\]
B) \[\frac{4(x-1)}{{{(1+x)}^{3}}}\]
C) \[\frac{x-1}{{{(1+x)}^{3}}}\]
D) \[\frac{4}{{{(x+1)}^{3}}}\]
Correct Answer: B
Solution :
Given,\[\frac{x}{1}=\frac{1-\sqrt{y}}{1+\sqrt{y}}\] Applying componendo and dividendo, we get \[\frac{1+x}{1-x}=\frac{(1+\sqrt{y})+(1-\sqrt{y})}{(1+\sqrt{y})-(1-\sqrt{y})}\] \[\Rightarrow \] \[\frac{1+x}{1-x}=\frac{2}{2\sqrt{y}}\] \[\Rightarrow \] \[y={{\left( \frac{1-x}{1+x} \right)}^{2}}\] On differentiating w.r.t.\[x\], we get \[\frac{dy}{dx}=\frac{-2{{(1+x)}^{2}}(1-x)-{{(1-x)}^{2}}\cdot 2(1+x)}{{{(1+x)}^{4}}}\] \[=\frac{(1-x)(1+x)(-2-2x-2+2x)}{{{(1+x)}^{4}}}\] \[=\frac{4(x-1)}{{{(x+1)}^{3}}}\]
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