# How do you find dy/dx by implicit differentiation of x^(1/2)+y^(1/2)=9?

Jan 11, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 - 9 {x}^{- \frac{1}{2}}$

#### Explanation:

Differentiate the equation with respect to $x$, considering that:

$\frac{d}{\mathrm{dx}} f \left(y \left(x\right)\right) = \frac{\mathrm{df}}{\mathrm{dy}} \cdot y ' \left(x\right)$

so:

$\frac{d}{\mathrm{dx}} \left({x}^{\frac{1}{2}} + {y}^{\frac{1}{2}}\right) = \frac{d}{\mathrm{dx}} \left(9\right)$

$\frac{1}{2} {x}^{- \frac{1}{2}} + \frac{1}{2} {y}^{- \frac{1}{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

From the original equation we see that:

${y}^{\frac{1}{2}} = 9 - {x}^{\frac{1}{2}}$

and we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{- \frac{1}{2}} \left({x}^{\frac{1}{2}} - 9\right) = 1 - 9 {x}^{- \frac{1}{2}}$