A) \[\frac{1}{2\,\,(1+{{x}^{2}})}\]
B) \[\frac{1}{1+{{x}^{2}}}\]
C) 1
D) - 1
Correct Answer: D
Solution :
\[\frac{d}{dx}{{\tan }^{-1}}\left[ \frac{\cos x-\sin x}{\cos x+\sin x} \right]\]\[=\frac{d}{dx}{{\tan }^{-1}}\left[ \tan \left( \frac{\pi }{4}-x \right) \right]=-1\].
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