Question

For the curve \( \sqrt{x} + \sqrt{y} = 1 \), find the value of \( \frac{dy}{dx} \) at the point \( \left(\frac{1}{9}, \frac{1}{9}\right) \).

Answer 1

Given: \[ \sqrt{x} + \sqrt{y} = 1 \] Rewriting in exponential form: \[ x^{1/2} + y^{1/2} = 1 \] Differentiating both sides with respect to \( x \): \[ \frac{d}{dx}\left(x^{1/2}\right) + \frac{d}{dx}\left(y^{1/2}\right) = \frac{d}{dx}(1) \] \[ \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0 \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = - \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{y}}} = - \frac{\sqrt{y}}{\sqrt{x}} \] Substituting the given point \( \left(\frac{1}{9}, \frac{1}{9}\right) \): \[ \sqrt{x} = \sqrt{\frac{1}{9}} = \frac{1}{3}, \quad \sqrt{y} = \sqrt{\frac{1}{9}} = \frac{1}{3} \] \[ \frac{dy}{dx} = - \frac{1/3}{1/3} = \boxed{-1} \]