MCQ
If $y=x \log x$ then value of $\frac{d^2 y}{d x^2}$ :
  • A
    $\frac{1}{1+x}$
  • $\frac{1}{x}$
  • C
    $\log (1+x)$
  • D
    $1+\log x$

Answer

Correct option: B.
$\frac{1}{x}$
(B)
$\begin{aligned} & & y & =x \log x \\ \Rightarrow & & \frac{d y}{d x} & =x \cdot \frac{1}{x}+\log x \cdot 1 \\ \Rightarrow & & \frac{d y}{d x} & =1+\log x \\ \Rightarrow & & \frac{d^2 y}{d x^2} & =0+\frac{1}{x}=\frac{1}{x}\end{aligned}$

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