Question

Find y' by implicit differentiation √x +√y =1

Answer 1

Differentiate both sides with respect to x

$\frac{d}{dx}(x^{\frac{1}{2}} + y^{\frac{1}{2}}) = \frac{d}{dx}(1)\\\\ \frac{1}{2x^{\frac{1}{2}}} + \frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = 0\\\\\frac{1}{2y^{\frac{1}{2}}}\frac{dy}{dx} = -\frac{1}{2x^{\frac{1}{2}}}\\\\\frac{1}{2\sqrt{y}}\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \\\\\frac{dy}{dx} = -\frac{1}{2\sqrt{x}} \times 2\sqrt{y}\\\\\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \\\\\boxed{\bf{y'= -\frac{\sqrt{y}}{\sqrt{x}}}}$