Answer
(a) $y' = -\frac{\sqrt{y}}{\sqrt{x}}$
(b) $y' = 1-\frac{1}{\sqrt{x}}$
(c) $y' = 1-\frac{1}{\sqrt{x}}$
Work Step by Step
(a) $\sqrt{x} + \sqrt{y} = 1$
$\frac{1}{2}x^{-1/2}+\frac{1}{2}y^{-1/2}~y' = 0$
$\frac{1}{2}y^{-1/2}~y' = -\frac{1}{2}x^{-1/2}$
$\frac{y'}{2\sqrt{y}} = -\frac{1}{2~\sqrt{x}}$
$y' = -\frac{\sqrt{y}}{\sqrt{x}}$
(b) $\sqrt{x} + \sqrt{y} = 1$
$\sqrt{y} = 1 - \sqrt{x}$
$y = 1 - 2\sqrt{x}+x$
$y' = -\frac{1}{\sqrt{x}}+1$
$y' = 1-\frac{1}{\sqrt{x}}$
(c) $y' = -\frac{\sqrt{y}}{\sqrt{x}}$
$y' = -\frac{\sqrt{1 - 2\sqrt{x}+x}}{\sqrt{x}}$
$y' = -\frac{\sqrt{(1 - \sqrt{x})^2}}{\sqrt{x}}$
$y' = -\frac{(1 - \sqrt{x})}{\sqrt{x}}$
$y' = 1-\frac{1}{\sqrt{x}}$
