A) \[\frac{1}{1+{{x}^{2}}}\]
B) \[-\frac{1}{1+{{x}^{2}}}\]
C) \[\frac{2}{1+{{x}^{2}}}\]
D) \[-\frac{2}{1+{{x}^{2}}}\]
Correct Answer: D
Solution :
Let \[y={{\cos }^{-1}}\frac{x-{{x}^{-1}}}{x+{{x}^{-1}}}\] On differentiating both sides w.r.t. \[x,\]we get \[\frac{dy}{dx}=-\frac{1}{\sqrt{1-{{\left( \frac{x-{{x}^{-1}}}{x+{{x}^{-1}}} \right)}^{2}}}}\frac{d}{dx}\left( \frac{x-{{x}^{-1}}}{x+{{x}^{-1}}} \right)\] \[=-\frac{x+{{x}^{-1}}}{\sqrt{{{(x+{{x}^{-1}})}^{2}}-{{(x-{{x}^{-1}})}^{2}}}}\] \[\times \,\,\frac{(x+{{x}^{-1}})(1+{{x}^{-2}})-(x-{{x}^{-1}})(1-{{x}^{-2}})}{{{(x+{{x}^{-1}})}^{2}}}\] \[=-\frac{1}{\sqrt{4}}\left( \frac{\begin{align} & x+{{x}^{-1}}+{{x}^{-1}}+{{x}^{-3}} \\ & -(x-{{x}^{-1}}-{{x}^{-1}}+{{x}^{-3}}) \\ \end{align}}{(x+{{x}^{-1}})} \right)\] \[=-\frac{1}{2}\frac{(4{{x}^{-1}})}{(x+{{x}^{-1}})}=-\frac{2}{({{x}^{2}}+1)}\] Alternate Solution: Let \[y={{\cos }^{-1}}\left( \frac{x-{{x}^{-1}}}{x+{{x}^{-1}}} \right)\] \[={{\cos }^{-1}}\left( \frac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)\] Put \[x=\cot \theta \] \[\therefore \] \[y={{\cos }^{-1}}\left( \frac{{{\cot }^{2}}\theta -1}{{{\cot }^{2}}\theta +1} \right)\] \[={{\cos }^{-1}}\left( \frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)\] \[={{\cos }^{-1}}\cos 2\theta =2\theta \] \[\Rightarrow \] \[y=2{{\cot }^{-1}}x\] On differentiating both sides w. r. t. \[x,\]we get \[\frac{dy}{dx}=-\frac{2}{1+{{x}^{2}}}\]
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