How do you determine dy/dx given x^(1/3)+y^(1/3)=7?

May 13, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(\frac{y}{x}\right)}^{\frac{2}{3}}$

Explanation:

Note that $\frac{d}{\mathrm{dx}} {x}^{\frac{1}{3}} = \frac{1}{3} {x}^{- \frac{2}{3}}$ through the power rule.

The derivative with respect to $x$ of ${y}^{\frac{1}{3}}$ is found identically, except for that since $y$ is its own function, the chain rule will come into effect, and we will have to multiply the derivative found through the product rule by the derivative of $y$, which is $\frac{\mathrm{dy}}{\mathrm{dx}}$.

That is $\frac{d}{\mathrm{dx}} {y}^{\frac{1}{3}} = \frac{1}{3} {y}^{- \frac{2}{3}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$.

So we have:

${x}^{\frac{1}{3}} + {y}^{\frac{1}{3}} = 7$

And taking the derivative of both sides:

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

Multiplying both sides by $3$ and rewriting:

$\frac{1}{x} ^ \left(\frac{2}{3}\right) + \frac{1}{y} ^ \left(\frac{2}{3}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

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

$\frac{1}{y} ^ \left(\frac{2}{3}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{x} ^ \left(\frac{2}{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{\frac{2}{3}} / {x}^{\frac{2}{3}} = - {\left(\frac{y}{x}\right)}^{\frac{2}{3}}$