A) \[\frac{y}{1-{{x}^{2}}}\]
B) \[\frac{y}{{{x}^{2}}-1}\]
C) \[\frac{y}{{{x}^{2}}+1}\]
D) None of these
Correct Answer: A
Solution :
\[y=\sqrt{\frac{1+x}{1-x}}\] On squaring both sides, we get \[{{y}^{2}}=\frac{1+x}{1-x}\] ...(i) On differentiating w.r.t.\[x,\] \[2y\frac{dy}{dx}=\frac{(1-x)(1)-(1+x)(-1)}{{{(1-x)}^{2}}}\] \[\Rightarrow \] \[2y\frac{dy}{dx}=\frac{1-x+1+x}{{{(1-x)}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{2y}.\frac{2}{{{(1-x)}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{y}{{{y}^{2}}{{(1-x)}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{y}{\left( \frac{1+x}{1-x} \right){{(1-x)}^{2}}}\] [from Eq. (i)] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{y}{(1+x)(1-x)}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{y}{1-{{x}^{2}}}\]
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