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If \(y=x\cdot {\log }_{e}x\), then the value of \(\frac{{d}^{2}y}{d{x}^{2}}\) will be -

Q1

If \(y=x\cdot {\log }_{e}x\), then the value of \(\frac{{d}^{2}y}{d{x}^{2}}\) will be -

a

\(\frac{1}{1+x}\)

b

\(\frac{1}{x}\)

c

\({\log }_{e}(1+x)\)

d

\(1+{\log }_{e}x\)

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