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