If \[f(x)=\sqrt{1+{{\cos }^{2}}({{x}^{2}})}\], then \[f'\left( \frac{\sqrt{\pi }}{2} \right)\] is [Orissa JEE 2004]

If \[f(x)=\sqrt{1+{{\cos }^{2}}({{x}^{2}})}\], then \[f'\left( \frac{\

1.	If \[f(x)=\sqrt{1+{{\cos }^{2}}({{x}^{2}})}\], then \[f'\left( \frac{\sqrt{\pi }}{2} \right)\] is [Orissa JEE 2004]
A.	\[\sqrt{\pi }/6\]
B.	\[-\,\sqrt{(\pi /6)}\]
C.	\[1/\sqrt{6}\]
D.	\[\pi /\sqrt{6}\]
Answer» C. \[1/\sqrt{6}\]

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If \[f(x)=\sqrt{1+{{\cos }^{2}}({{x}^{2}})}\], then \[f'\left( \frac{\sqrt{\pi }}{2} \right)\] is [Orissa JEE 2004]

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