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