If \[y={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+{{\sec }^{-1}}\frac{1+{{x}^{2}}}{1-{{x}^{2}}}\], then \[\frac{dy}{dx}\]= [RPET 1996]

If \[y={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+{{\sec }^{-1}}\frac{1+{{x}

1.	If \[y={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+{{\sec }^{-1}}\frac{1+{{x}^{2}}}{1-{{x}^{2}}}\], then \[\frac{dy}{dx}\]= [RPET 1996]
A.	\[\frac{4}{1-{{x}^{2}}}\]
B.	\[\frac{1}{1+{{x}^{2}}}\]
C.	\[\frac{4}{1-{{x}^{2}}}\]
D.	\[\frac{-4}{1+{{x}^{2}}}\]
Answer» D. \[\frac{-4}{1+{{x}^{2}}}\]

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If \[y={{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}+{{\sec }^{-1}}\frac{1+{{x}^{2}}}{1-{{x}^{2}}}\], then \[\frac{dy}{dx}\]= [RPET 1996]

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