If \[y={{x}^{2}}\log x+\frac{2}{\sqrt{x}},\] then \[\frac{dy}{dx}=\]

1.	If \[y={{x}^{2}}\log x+\frac{2}{\sqrt{x}},\] then \[\frac{dy}{dx}=\]
A.	\[x+2x\log x-\frac{1}{\sqrt{x}}\]
B.	\[x+2x\log x-\frac{1}{{{x}^{3/2}}}\]
C.	\[x+2x\log x-\frac{2}{{{x}^{3/2}}}\]
D.	None of these
Answer» C. \[x+2x\log x-\frac{2}{{{x}^{3/2}}}\]

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If \[y={{x}^{2}}\log x+\frac{2}{\sqrt{x}},\] then \[\frac{dy}{dx}=\]

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