# How do you find the second derivative by implicit differentiation on x^3y^3=8 ?

Jul 31, 2014

As the first step, we will differentiate once, and apply the product rule:

$\frac{d}{\mathrm{dx}} \left[{x}^{3}\right] \cdot {y}^{3} + \frac{d}{\mathrm{dx}} \left[{y}^{3}\right] \cdot {x}^{3} = \frac{d}{\mathrm{dx}} \left[8\right]$

For ${y}^{3}$, remember to use the chain rule. Simplifying yields:

$3 {x}^{2} {y}^{3} + 3 {y}^{2} {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Now, we will solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

We can cancel off the 3, an ${x}^{2}$, and a ${y}^{2}$, which will yield:

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

Now, differentiate once again. We will apply the quotient rule:

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

Looking back at the previous equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we can substitute into our equation for the second derivative to get it in terms of only $x$ and $y$:

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

Simplifying yields:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 y}{x} ^ 2$