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The second derivative of $y=x\, log x $ is : |
$1+log\, x $ $log \, x$ $x$ $\frac{1}{x}$ |
$\frac{1}{x}$ |
The correct answer is Option (4) → $\frac{1}{x}$ $y=x\, log x $ $⇒\frac{dy}{dx}=x×\frac{1}{x}+\log x=1+\log x$ $⇒\frac{d^2y}{dx^2}=0+\frac{1}{x}=\frac{1}{x}$ |
