If y = √(2024x+2025), then which of the following is correct?

The correct answer is Option (4) → $y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2=0$

Given

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

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

$\frac{dy}{dx}=\frac{1}{2}(2024x+2025)^{-\frac{1}{2}}\cdot2024$

$\frac{dy}{dx}=\frac{1012}{\sqrt{2024x+2025}}$

$\frac{d^2y}{dx^2}=1012\cdot\left(-\frac{1}{2}\right)(2024x+2025)^{-\frac{3}{2}}\cdot2024$

$\frac{d^2y}{dx^2}=-\frac{1012\cdot2024}{2(2024x+2025)^{\frac{3}{2}}}$

$\frac{d^2y}{dx^2}=-\frac{1012\cdot1012}{(2024x+2025)^{\frac{3}{2}}}$

$y=\sqrt{2024x+2025}$, $\frac{dy}{dx}=\frac{1012}{y}$

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

Now compute

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

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

$=-\frac{1012^2}{y^2}+\frac{1012^2}{y^2}$

$=0$

Hence the correct relation is $y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2=0$