A) -2
B) -1
C) 0
D) 1
Correct Answer: D
Solution :
Let \[y={{\cos }^{-1}}\,\left( \frac{{{x}^{-1}}-x}{{{x}^{-1}}+x} \right)\] \[y={{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)\] \[y=2\,{{\tan }^{-1}}\,x\] \[\frac{dy}{dx}=\frac{2}{1+{{x}^{2}}}\] At \[x=-1\] \[\frac{dy}{dx}=1\]You need to login to perform this action.
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