If $y = \sqrt{2024x + 2025}$, then which of the following is correct?

Given $y = \sqrt{2024x + 2025}$

$y = (2024x + 2025)^{1/2}$

Using the chain rule:

$\frac{dy}{dx} = \frac{1}{2}(2024x + 2025)^{-1/2} \times 2024$

$\frac{dy}{dx} = \frac{2024}{2\sqrt{2024x + 2025}}$

Since $y = \sqrt{2024x + 2025}$:

$\frac{dy}{dx} = \frac{1012}{y}$

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To find the second derivative, rewrite $\frac{dy}{dx} = 1012y^{-1}$

$\frac{d^2y}{dx^2} = 1012 \times (-1)y^{-2} \times \frac{dy}{dx}$

$\frac{d^2y}{dx^2} = -\frac{1012}{y^2} \times \frac{dy}{dx}$

Substituting $\frac{dy}{dx} = \frac{1012}{y}$:

$\frac{d^2y}{dx^2} = -\frac{1012}{y^2} \times \frac{1012}{y}$

$\frac{d^2y}{dx^2} = -\frac{1012^2}{y^3}$

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Consider the equation: $y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$

The first term:

$y\frac{d^2y}{dx^2} = y \times \left(-\frac{1012^2}{y^3}\right)$

$y\frac{d^2y}{dx^2} = -\frac{1012^2}{y^2}$

The second term:

$\left(\frac{dy}{dx}\right)^2 = \left(\frac{1012}{y}\right)^2$

$\left(\frac{dy}{dx}\right)^2 = \frac{1012^2}{y^2}$

Therefore:

$y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = -\frac{1012^2}{y^2} + \frac{1012^2}{y^2}$

$y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$

The equation $y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = 0$ is satisfied.