A) \[{{n}^{2}}y\]
B) \[-{{n}^{2}}y\]
C) \[-y\]
D) \[2{{x}^{2}}y\]
Correct Answer: A
Solution :
Given,\[y={{(x+\sqrt{1+{{x}^{2}}})}^{n}}\] On differentiating w. r. t. x, we get \[\frac{dy}{dx}=n{{(x+\sqrt{1+{{x}^{2}}})}^{n-1}}\left( 1+\frac{x}{\sqrt{1+{{x}^{2}}}} \right)\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{n{{(x+\sqrt{1+{{x}^{2}}})}^{n}}}{\sqrt{1+{{x}^{2}}}}\] \[\Rightarrow \] \[\sqrt{1+{{x}^{2}}}\frac{dy}{dx}=n{{(x+\sqrt{1+{{x}^{2}}})}^{n}}\] Again differentiating w.r.t. x, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\cdot \sqrt{1+{{x}^{2}}}+\frac{dy}{dx}\left( \frac{x}{\sqrt{1+{{x}^{2}}}} \right)\] \[={{n}^{2}}{{(x+\sqrt{1+{{x}^{2}}})}^{n-1}}\left( 1+\frac{x}{\sqrt{1+{{x}^{2}}}} \right)\] \[\Rightarrow \] \[(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\cdot \frac{dy}{dx}={{n}^{2}}{{(x+\sqrt{1+{{x}^{2}}})}^{n}}\] \[\Rightarrow \] \[(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}={{n}^{2}}y\]
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