# How do you find the derivative of sqrtx+sqrty=9?

Apr 13, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sqrt{\frac{y}{x}}$

#### Explanation:

$\sqrt{x} + \sqrt{y} = 9$

By rewriting a bit,

$R i g h t a r r o w {x}^{\frac{1}{2}} + {y}^{\frac{1}{2}} = 9$

By differentiating with respect to $x$,

$R i g h t a r r o w \frac{1}{2} {x}^{- \frac{1}{2}} + \frac{1}{2} {y}^{- \frac{1}{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

By cleaning up a bit,

$R i g h t a r r o w \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

By subtracting $\frac{1}{2 \sqrt{x}}$ from both sides,

$\frac{1}{2 \sqrt{y}} \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \sqrt{x}}$

By multiplying both sides by $2 \sqrt{y}$,

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\cancel{2} \sqrt{y}}{\cancel{2} \sqrt{x}} = - \setminus \sqrt{\frac{y}{x}}$

I hope that this was clear.