# How do you determine dy/dx given sqrtx+sqrty=25?

Aug 25, 2016

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

#### Explanation:

$\sqrt{x} + \sqrt{y} = 25$.

$\therefore \frac{d}{\mathrm{dx}} \left(\sqrt{x} + \sqrt{y}\right) = \frac{d}{\mathrm{dx}} 25 = 0$.

$\therefore \frac{d}{\mathrm{dx}} \left(\sqrt{x}\right) + \frac{d}{\mathrm{dx}} \left(\sqrt{y}\right) = 0$.

$\therefore \frac{1}{2 \sqrt{x}} + \frac{d}{\mathrm{dy}} \left(\sqrt{y}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0. \ldots \ldots \ldots . . \text{.[Chain Rule]}$.

$\therefore \frac{1}{2 \sqrt{x}} + \frac{1}{2 \sqrt{y}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{2 \sqrt{x}} / \left(\frac{1}{2 \sqrt{y}}\right)$

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

Aug 25, 2016

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

#### Explanation:

Here you need to use implicit differentiation because your $y$ is difficult to be left alone on one side as in our usual functons; but it is not a problem, only remember that $y$ is itself a function of $x$ and when you differentiate it you need to include the term $\frac{\mathrm{dy}}{\mathrm{dx}}$ to take into account this dependence; the rest is as usual.

So let us differentiate:

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

you see the appearance of the term $\frac{\mathrm{dy}}{\mathrm{dx}}$ after differentiating $\sqrt{y}$!!!

rearrange:

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