# How do you twice differentiate x^3+y^3=1?

Jun 30, 2015

$y ' ' = - \frac{2 x}{{y}^{5}}$

#### Explanation:

${x}^{3} + {y}^{3} = 1$

We use implicit differentiation:

$D \left[{x}^{3} + {y}^{3}\right] = D \left[1\right]$

So:

$3 {x}^{2} + 3 {y}^{2} y ' = 0$

y'=-(cancel(3)x^2)/(cancel(3)y^2)=-(x^2)/(y^(2) $\textcolor{red}{\left(1\right)}$

So now we have to differentiate again using the quotient rule. A useful tip is to start and finish with the function on the bottom $\Rightarrow$

$y ' ' = \frac{{y}^{2} \left(- 2 x\right) - \left(- {x}^{2}\right) .2 y . y '}{{y}^{4}}$ $\textcolor{red}{\left(2\right)}$

We already know from $\textcolor{red}{\left(1\right)}$ that :

$y ' = - {x}^{2} / {y}^{2}$

So we can substitute that expression for $y '$ into $\textcolor{red}{\left(2\right)}$ $\Rightarrow$

y''=(y^(2)(-2x)-(-x^(2)).2y.((-x^(2))/(y^(2))))/(y^(4)

$y ' ' = \frac{- 2 x {y}^{2} - \frac{2 {x}^{4}}{y}}{{y}^{4}}$

Multiplying top and bottom by $y$ $\Rightarrow$

$y ' ' = \frac{\left[- 2 x {y}^{3} - 2 {x}^{4}\right]}{{y}^{5}}$

$y ' ' = \frac{- 2 x \left[{y}^{3} + {x}^{3}\right]}{{y}^{5}}$ $\textcolor{red}{\left(3\right)}$

Since we know that:

${x}^{3} + {y}^{3} = 1$

We can substitute that value of $1$ into $\textcolor{red}{\left(3\right)} \Rightarrow$

$y ' ' = \frac{- 2 x}{{y}^{5}}$