If \[y={{e}^{{{\tan }^{-1}}x}}\], then \[(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=\]

If \[y={{e}^{{{\tan }^{-1}}x}}\], then \[(1+{{x}^{2}})\frac{{{d}^{2}}y

1.	If \[y={{e}^{{{\tan }^{-1}}x}}\], then \[(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=\]
A.	\[(1-2x)\frac{dy}{dx}\]
B.	\[-2x\frac{dy}{dx}\]
C.	\[-x\frac{dy}{dx}\]
D.	0
Answer» B. \[-2x\frac{dy}{dx}\]

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If \[y={{e}^{{{\tan }^{-1}}x}}\], then \[(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}=\]

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