# How do you determine dy/dx given x^(1/4)+y^(1/4)=2?

Oct 28, 2016

Well, the hardest part is remembering that $\frac{\mathrm{dy}}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$.

If you've got this in your head, finding dy/dx should be fairly easy...

To simplify matters though, I will turn 1/4 into the variable n...

This means that:

${x}^{n} + {y}^{n} = 2$

Also...

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

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

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

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

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

Now, since n=1/4:

$n - 1$

$= \frac{1}{4} - \frac{4}{4}$

$= - \frac{3}{4}$

Which means that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(\frac{x}{y}\right)}^{- \frac{3}{4}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {\left(\frac{y}{x}\right)}^{\frac{3}{4}}$

This can be made to look prettier...

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sqrt[4]{{\left(\frac{y}{x}\right)}^{3}}$

And presto!!