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

Given $y = \sqrt{ax + b}$, find $y\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2$

Rewrite as $y = (ax + b)^{1/2}$

$\frac{dy}{dx} = \frac{1}{2}(ax + b)^{-1/2} \cdot a$

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

Since $y = \sqrt{ax + b}$:

$\frac{dy}{dx} = \frac{a}{2y}$

Starting from $\frac{dy}{dx} = \frac{a}{2}(ax + b)^{-1/2}$:

$\frac{d^2y}{dx^2} = \frac{a}{2} \cdot \left(-\frac{1}{2}\right)(ax + b)^{-3/2} \cdot a$

$\frac{d^2y}{dx^2} = -\frac{a^2}{4(ax + b)^{3/2}}$

Since $y = \sqrt{ax + b}$, we have $(ax + b)^{3/2} = y^3$:

$\frac{d^2y}{dx^2} = -\frac{a^2}{4y^3}$

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

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

$= 0$

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