Q83 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\text { If } y=\log _e\left(\dfrac{x^2}{e^2}\right) \text {, then } \dfrac{d^2 y}{d x^2} \text { is equal to }$
- A$-\dfrac{1}{x^2}$
- B$-\dfrac{1}{x}$
- C$\dfrac{2}{x^2}$
- Correct answerD$-\dfrac{2}{x^2}$
Solution
Given $y = \log_{e}\left(\dfrac{x^2}{e^2}\right)$.
Using the properties of logarithms, $\log_{e}\left(\dfrac{a}{b}\right) = \log_{e} a - \log_{e} b$, we have:
$y = \log_{e}(x^2) - \log_{e}(e^2)$
$y = 2\log_{e} x - 2\log_{e} e$
Since $\log_{e} e = 1$, the expression simplifies to:
$y = 2\log_{e} x - 2$
Differentiating with respect to $x$ once:
$\dfrac{dy}{dx} = \dfrac{d}{dx}(2\log_{e} x - 2) = 2 \times \dfrac{1}{x} = \dfrac{2}{x}$
Differentiating with respect to $x$ again to find the second derivative:
$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{2}{x}\right) = 2 \times \left(-\dfrac{1}{x^2}\right) = -\dfrac{2}{x^2}$
Answer: $-\dfrac{2}{x^2}$
Using the properties of logarithms, $\log_{e}\left(\dfrac{a}{b}\right) = \log_{e} a - \log_{e} b$, we have:
$y = \log_{e}(x^2) - \log_{e}(e^2)$
$y = 2\log_{e} x - 2\log_{e} e$
Since $\log_{e} e = 1$, the expression simplifies to:
$y = 2\log_{e} x - 2$
Differentiating with respect to $x$ once:
$\dfrac{dy}{dx} = \dfrac{d}{dx}(2\log_{e} x - 2) = 2 \times \dfrac{1}{x} = \dfrac{2}{x}$
Differentiating with respect to $x$ again to find the second derivative:
$\dfrac{d^2 y}{dx^2} = \dfrac{d}{dx}\left(\dfrac{2}{x}\right) = 2 \times \left(-\dfrac{1}{x^2}\right) = -\dfrac{2}{x^2}$
Answer: $-\dfrac{2}{x^2}$
Hence, the correct answer is option: 'D'.