A) \[y\]
B) \[2y\]
C) \[0\]
D) \[-2y\]
Correct Answer: B
Solution :
\[y=\frac{{{x}^{2}}}{2}+\frac{1}{2}x\sqrt{{{x}^{2}}+1}+\ln \sqrt{x+\sqrt{{{x}^{2}}+1}}\] \[\Rightarrow \] \[y'=x+\frac{1}{2}\sqrt{{{x}^{2}}+1}+\frac{1}{2}x\frac{x}{\sqrt{{{x}^{2}}+1}}\] \[+\frac{1}{2(x+\sqrt{{{x}^{2}}+1})}\times \left( 1+\frac{x}{\sqrt{{{x}^{2}}+1}} \right)\] \[=x+\frac{({{x}^{2}}+1+{{x}^{2}})}{2\sqrt{{{x}^{2}}+1}}+\frac{1}{2\sqrt{{{x}^{2}}+1}}\] \[=x+\frac{2{{x}^{2}}+1}{2\sqrt{{{x}^{2}}+1}}+\frac{1}{2\sqrt{{{x}^{2}}+1}}\] \[=x+\frac{2{{x}^{2}}+1+1}{2\sqrt{{{x}^{2}}+1}}\] \[\Rightarrow \] \[y'=x+\sqrt{{{x}^{2}}+1}\] \[\Rightarrow \] \[xy'=logy'\] \[=x(x+\sqrt{{{x}^{2}}+1})+\log \,(x+\sqrt{{{x}^{2}}+1})\] \[={{x}^{2}}+x\sqrt{{{x}^{2}}+1}+\log \,(x+\sqrt{{{x}^{2}}+1})=2y\]
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