If $y = \log_e\left(\frac{e^2}{x^2}\right)$ for $x \neq 0$, then $\frac{d^2y}{dx^2}$ equals

Given: $y = \log_e\left(\frac{e^2}{x^2}\right)$

Using the quotient rule for logarithms:

$y = \log_e(e^2) - \log_e(x^2)$

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Using the power rule for logarithms:

$y = 2\log_e(e) - 2\log_e(x)$

Since $\log_e(e) = 1$:

$y = 2(1) - 2\ln(x)$

$y = 2 - 2\ln(x)$

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Finding the first derivative:

$\frac{dy}{dx} = \frac{d}{dx}(2) - \frac{d}{dx}(2\ln x)$

$\frac{dy}{dx} = 0 - 2 \cdot \frac{1}{x}$

$\frac{dy}{dx} = -\frac{2}{x}$

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Finding the second derivative:

Rewriting: $\frac{dy}{dx} = -2x^{-1}$

Using the power rule:

$\frac{d^2y}{dx^2} = -2 \cdot (-1)x^{-2}$

$\frac{d^2y}{dx^2} = 2x^{-2}$

$\frac{d^2y}{dx^2} = \frac{2}{x^2}$

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Therefore, $\frac{d^2y}{dx^2} = \frac{2}{x^2}$