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

Sep 10, 2014

By implicitly differentiating twice, we can find
$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{2 x}{y} ^ 5$.

First, let us find $\frac{\mathrm{dy}}{\mathrm{dx}}$.
${x}^{3} + {y}^{3} = 1$
by differentiating with respect to $x$,
$R i g h t a r r o w 3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
by subtracting $3 {x}^{2}$,
$R i g h t a r r o w 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2}$
by dividing by $3 {y}^{2}$,
$R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{x}^{2}}{{y}^{2}}$

Now, let us find $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$.
by differentiating with respect to $x$,
Rightarrow{d^2y}/{dx^2}=-{2x cdot y^2-x^2 cdot 2y{dy}/{dx}}/{(y^2)^2} =-{2x(y^2-xy{dy}/{dx})}/{y^4}
by plugging in $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{x}^{2}}{{y}^{2}}$,
$R i g h t a r r o w \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{2 x \left[{y}^{2} - x y \left(- {x}^{2} / {y}^{2}\right)\right]}{y} ^ 4 = - \frac{2 x \left({y}^{2} + {x}^{3} / y\right)}{y} ^ 4$
by multiplying the numerator and the denominator by $y$,
$R i g h t a r r o w \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{2 x \left({y}^{3} + {x}^{3}\right)}{y} ^ 5$
by plugging in ${y}^{3} + {x}^{3} = 1$,
$R i g h t a r r o w \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{2 x}{y} ^ 5$