A) \[\frac{-2}{(1+{{x}^{2}})},x\ne 0\]
B) \[\frac{2}{(1+{{x}^{2}})},x\ne 0\]
C) \[\frac{2(1-{{x}^{2}})}{(1+{{x}^{2}})|1-{{x}^{2}}|},x\ne \pm 1,0\]
D) None of the above
Correct Answer: C
Solution :
\[\frac{d}{dx}\cos e{{c}^{-1}}\left( \frac{1+{{x}^{2}}}{2x} \right)=\frac{d}{dx}{{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right)\] \[=\frac{1}{\sqrt{1-{{\left( \frac{2x}{1+{{x}^{2}}} \right)}^{2}}}}\left( \frac{(1+{{x}^{2}})2-(2x)(2x)}{{{(1+{{x}^{2}})}^{2}}} \right)\] \[=\frac{2(1-{{x}^{2}})}{(1+{{x}^{2}})|1-{{x}^{2}}|},x\ne \pm 1,0\]
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