If $y =\sqrt{ax + b}$, then $y\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2=$

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The correct answer is Option (1) → 0

Given:

$y = \sqrt{ax + b}$

First derivative:

$\frac{dy}{dx} = \frac{a}{2\sqrt{ax + b}} = \frac{a}{2y}$

Second derivative:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{a}{2y}\right) = -\frac{a}{2y^2} \cdot \frac{dy}{dx} = -\frac{a}{2y^2} \cdot \frac{a}{2y} = -\frac{a^2}{4y^3}$

Now, compute:

$y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 = y\left(-\frac{a^2}{4y^3}\right) + \left(\frac{a}{2y}\right)^2$

$= -\frac{a^2}{4y^2} + \frac{a^2}{4y^2} = 0$

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