A) \[(1-2x)\frac{dy}{dx}\]
B) \[-2x\frac{dy}{dx}\]
C) \[-x\frac{dy}{dx}\]
D) 0
Correct Answer: A
Solution :
\[y={{e}^{{{\tan }^{-1}}x}}\Rightarrow \frac{dy}{dx}=\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}\] \[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{(1+{{x}^{2}}).\frac{{{e}^{{{\tan }^{-1}}x}}}{(1+{{x}^{2}})}-{{e}^{{{\tan }^{-1}}x}}(2x)}{{{(1+{{x}^{2}})}^{2}}}\] \[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{(1-2x){{e}^{{{\tan }^{-1}}x}}}{{{(1+{{x}^{2}})}^{2}}}\]\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}(1+{{x}^{2}})=(1-2x)\frac{dy}{dx}\].
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