A) \[{{n}^{2}}y\]
B) \[-{{n}^{2}}y\]
C) \[-y\]
D) \[2{{x}^{2}}y\]
Correct Answer: A
Solution :
\[y={{(x+\sqrt{1+{{x}^{2}}})}^{n}}\Rightarrow \frac{dy}{dx}=n{{(x+\sqrt{1+{{x}^{2}}})}^{n-1}}\left( 1+\frac{x}{\sqrt{1+{{x}^{2}}}} \right)\] Þ \[\frac{dy}{dx}=\frac{n{{(x+\sqrt{1+{{x}^{2}}})}^{n}}}{\sqrt{1+{{x}^{2}}}}\] Þ \[(\sqrt{1+{{x}^{2}}})\,\frac{dy}{dx}=n{{(x+\sqrt{1+{{x}^{2}}})}^{n}}\] Þ \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\,.\,\sqrt{1+{{x}^{2}}}+\frac{dy}{dx}\left( \frac{x}{\sqrt{1+{{x}^{2}}}} \right)\] \[={{n}^{2}}{{\left( x+\sqrt{1+{{x}^{2}}} \right)}^{n-1}}\,\left( 1+\frac{x}{\sqrt{1+{{x}^{2}}}} \right)\] Þ \[(1+{{x}^{2}})\,.\,\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\,.\,\frac{dy}{dx}\,={{n}^{2}}{{(x+\sqrt{1+{{x}^{2}}})}^{n}}\] Þ \[(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\,.\,\frac{dy}{dx}={{n}^{2}}y\].
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