# How do you find y'' by implicit differentiation of x^3+y^3=1 ?

Sep 6, 2014

Differentiate one step at a time, then when you have a $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2$ term (i.e. y'') or can easily get one, rearrange the equation.

Implicit differentiation is remarkably similar to "regular" differentiation. We just need to treat any term with a y in it slightly differently.

First, we differentiate both sides of the equation:
$\frac{d}{\mathrm{dx}} \left({x}^{3} + {y}^{3}\right) = \frac{d}{\mathrm{dx}} \left(1\right)$

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

We know that $\frac{d}{\mathrm{dx}} {x}^{3} = 3 {x}^{2}$ and $\frac{d}{\mathrm{dx}} 1 = 0$, so
$3 {x}^{2} + \frac{d}{\mathrm{dx}} \left({y}^{3}\right) = 0$

Now we use the chain rule to implicitly find $\frac{d}{\mathrm{dx}} \left({y}^{3}\right)$:
Let $u = {y}^{3}$
Then $\frac{\mathrm{du}}{\mathrm{dy}} = 3 {y}^{2}$
and since $\frac{\mathrm{du}}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{du}}{\mathrm{dx}}$
we have $\frac{d}{\mathrm{dx}} \left({y}^{3}\right) = 3 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$.

So we have $3 {x}^{2} + 3 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 0$, and we can rearrange our equation to get $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 {x}^{2}}{3 {y}^{2}}$.

From here, the next step is to again differentiate both sides, this time using the quotient rule:
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(- \frac{3 {x}^{2}}{3 {y}^{2}}\right) = - \frac{d}{\mathrm{dx}} \left(\frac{3 {x}^{2}}{3 {y}^{2}}\right)$
$= \frac{3 {y}^{2} \cdot \frac{d}{\mathrm{dx}} \left(3 {x}^{2}\right) - 3 {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left(3 {y}^{2}\right)}{3 {y}^{2}} ^ 2$
$= - \frac{6 x 3 {y}^{2} - 3 {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left(3 {y}^{2}\right)}{9 {y}^{4}}$

To find $\frac{d}{\mathrm{dx}} \left(3 {y}^{2}\right)$, we again implicitly differentiate to get $6 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

And so:
$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{18 x {y}^{2} - 18 {x}^{2} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}}{9 {y}^{4}}$

Finally, recall that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 {x}^{2}}{3 {y}^{2}}$

So: $\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \frac{18 x {y}^{2} - 18 {x}^{2} y \cdot \frac{- 3 {x}^{2}}{3 {y}^{2}}}{9 {y}^{4}}$.