# How do you determine dy/dx given ysqrtx+xsqrty=16?

##### 1 Answer
Oct 7, 2016

$y ' = \frac{- \frac{y}{2 \sqrt{x}} - \sqrt{y}}{\sqrt{x} + \frac{x}{2 \sqrt{y}}}$

#### Explanation:

We have to use implicit differentiation which is basically just a special case of the chain rule. Every time we differentiate a factor that has a $y$ variable we have to include the factor $\frac{\mathrm{dy}}{\mathrm{dx}}$ or $y '$.

We will also be using the product rule and power rule to answer this question.

Begin by changing all of the radicals to their exponent equivalents.

$y {x}^{\frac{1}{2}} + x {y}^{\frac{1}{2}} = 16$

$y \left(\frac{1}{2}\right) {x}^{- \frac{1}{2}} + \left(1\right) y ' {x}^{\frac{1}{2}} + x \left(\frac{1}{2}\right) {y}^{- \frac{1}{2}} y ' + \left(1\right) {y}^{\frac{1}{2}} = 0$

Simplify and switch back to radicals

$\frac{y}{2 \sqrt{x}} + y ' \sqrt{x} + \frac{x y '}{2 \sqrt{y}} + \sqrt{y} = 0$

Gather terms with $y '$ to one side and everything else to the other side

$y ' \sqrt{x} + \frac{x y '}{2 \sqrt{y}} = - \frac{y}{2 \sqrt{x}} - \sqrt{y}$

Factor out $y '$

$y ' \left(\sqrt{x} + \frac{x}{2 \sqrt{y}}\right) = - \frac{y}{2 \sqrt{x}} - \sqrt{y}$

Isolate $y '$ by dividing

$y ' \frac{\cancel{\sqrt{x} + \frac{x}{2 \sqrt{y}}}}{\cancel{\sqrt{x} + \frac{x}{2 \sqrt{y}}}} = \frac{- \frac{y}{2 \sqrt{x}} - \sqrt{y}}{\sqrt{x} + \frac{x}{2 \sqrt{y}}}$

$y ' = \frac{- \frac{y}{2 \sqrt{x}} - \sqrt{y}}{\sqrt{x} + \frac{x}{2 \sqrt{y}}}$

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