How do you use implicit differentiation to find (dy)/(dx) given x^(2/3)+y^(2/3)=1?

Aug 29, 2016

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

Explanation:

Use the power rule and implicit differentiation to find the derivative.

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

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

Since we're differentiating with respect to $x$, we must put $\frac{\mathrm{dy}}{\mathrm{dx}}$ at every y term that we differentiate (after differentiating, of course).

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

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

Isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$ and simplify:

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 6 {y}^{\frac{1}{3}}}{6 {x}^{\frac{1}{3}}}$

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

Hopefully this helps!

Aug 29, 2016

$- {y}^{\frac{1}{3}} / {x}^{\frac{1}{3}}$

Explanation:

Define $f \left(x , y\right) = {x}^{\frac{2}{3}} + {y}^{\frac{2}{3}} - 1 = 0$

we know that

$\mathrm{df} = {f}_{x} \mathrm{dx} + {f}_{y} \mathrm{dy} = 0$

so

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