# How do you determine dy/dx given x^(2/3)+y^(2/3)=a^(2/3)?

Nov 14, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(\frac{y}{x}\right)}^{\frac{1}{3}}$

#### Explanation:

We will differentiate implicitly. This means, that whenever we differentiate a variable that's not $x$, the chain rule will kick in. Take a look at a couple examples before starting here:

• $\frac{d}{\mathrm{dx}} {x}^{3} = 3 {x}^{2} \text{ }$ but $\text{ } \frac{d}{\mathrm{dx}} {y}^{3} = 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$
• $\frac{d}{\mathrm{dx}} x = 1 \text{ }$ but $\text{ } \frac{d}{\mathrm{dx}} y = \frac{\mathrm{dy}}{\mathrm{dx}}$
• $\frac{d}{\mathrm{dx}} x {y}^{2} = \left(\frac{d}{\mathrm{dx}} x\right) {y}^{2} + x \left(\frac{d}{\mathrm{dx}} {y}^{2}\right) = {y}^{2} + x \left(2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

So, for the given example, we see that we will use the power rule on the left-hand side. On the right, don't be misled by the two-thirds power: ${a}^{\frac{2}{3}}$ is still a constant.

$\frac{d}{\mathrm{dx}} \left({x}^{\frac{2}{3}} + {y}^{\frac{2}{3}}\right) = \frac{d}{\mathrm{dx}} {a}^{\frac{2}{3}}$

Power rule:

$\frac{2}{3} {x}^{- \frac{1}{3}} + \frac{2}{3} {y}^{- \frac{1}{3}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Multiply both sides by $\frac{3}{2}$:

${x}^{- \frac{1}{3}} + {y}^{- \frac{1}{3}} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{1}{x} ^ \left(\frac{1}{3}\right) + \frac{1}{y} ^ \left(\frac{1}{3}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{1}{y} ^ \left(\frac{1}{3}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x} ^ \left(\frac{1}{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{\frac{1}{3}} / {x}^{\frac{1}{3}} = - {\left(\frac{y}{x}\right)}^{\frac{1}{3}} = - \sqrt[3]{\frac{y}{x}}$

Nov 14, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\sqrt[2]{\left({a}^{\frac{2}{3}} - {x}^{\frac{2}{3}}\right)}}{\sqrt[3]{x}}$

#### Explanation:

Rewrite it ${y}^{\frac{2}{3}} = {a}^{\frac{2}{3}} - {x}^{\frac{2}{3}}$ so that we have
$y = \sqrt[2]{{\left({a}^{\frac{2}{3}} - {x}^{\frac{2}{3}}\right)}^{3}}$ and the derivative is

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2} \sqrt[2]{\left({a}^{\frac{2}{3}} - {x}^{\frac{2}{3}}\right)} \cdot \left(- \frac{2}{3} {x}^{- \frac{1}{3}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\sqrt[2]{\left({a}^{\frac{2}{3}} - {x}^{\frac{2}{3}}\right)}}{\sqrt[3]{x}}$