If y=x _e x then value of d^2 y d x^2 : - Vidyadip

MCQ

If $y=x \log _e x$ then value of $\frac{d^2 y}{d x^2}$ :

A
$\frac{1}{1+x}$
B
$1+\log _e x$
C
$\log _e(1+x)$
✓
$\frac{1}{x}$

✓

Answer

Correct option: D.

$\frac{1}{x}$

(D)Given that $y=x \log _e x$
$\therefore \quad \frac{d y}{d x}=1 \cdot \log x+x \cdot \frac{1}{x}=1+\log _e x$
again differentiating
$\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(1+\log _c x\right)=0+\frac{1}{x}=\frac{1}{x}$
$\Rightarrow \quad \frac{d^2 y}{d x^2}=\frac{1}{x}$
Hence correct option is (D).

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