# How do you find the second derivative of x^3 + y^3 = 6xy - 1?

Jul 24, 2015

$\frac{{d}^{2} y}{\mathrm{dx}} ^ \left\{2\right\} = \frac{12 {x}^{2} {y}^{2} - 16 x y - 2 {x}^{4} y - 2 x {y}^{4}}{{\left({y}^{2} - 2 x\right)}^{3}}$

#### Explanation:

Assuming the equation ${x}^{3} + {y}^{3} = 6 x y - 1$ implicitly defines $y$ as a function of $x$, we can differentiate both sides with respect to $x$, using the Power Rule, Chain Rule, and Product Rule to get:

$3 {x}^{2} + 3 {y}^{2} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 6 y + 6 x \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

Now solve this equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$ as follows:

$\left(3 {y}^{2} - 6 x\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 6 y - 3 {x}^{2}$

$\setminus R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 y - 3 {x}^{2}}{3 {y}^{2} - 6 x} = \frac{2 y - {x}^{2}}{{y}^{2} - 2 x}$

Now differentiate this equation with respect to $x$ using all the rules mentioned, as well as the Quotient Rule:

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

$= \frac{2 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x {y}^{2} - 4 x \frac{\mathrm{dy}}{\mathrm{dx}} + 4 {x}^{2} - 4 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 4 y + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {x}^{2}}{{\left({y}^{2} - 2 x\right)}^{2}}$

$= \frac{\left(2 {x}^{2} y - 4 x - 2 {y}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {x}^{2} - 2 x {y}^{2} + 4 y}{{\left({y}^{2} - 2 x\right)}^{2}}$

Now substitute $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 y - {x}^{2}}{{y}^{2} - 2 x}$ into this last equation to get:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ \left\{2\right\} = \frac{\left(2 {x}^{2} y - 4 x - 2 {y}^{2}\right) \frac{2 y - {x}^{2}}{{y}^{2} - 2 x} + 2 {x}^{2} - 2 x {y}^{2} + 4 y}{{\left({y}^{2} - 2 x\right)}^{2}}$

$= \frac{\left(2 {x}^{2} y - 4 x - 2 {y}^{2}\right) \left(2 y - {x}^{2}\right) + \left(2 {x}^{2} - 2 x {y}^{2} + 4 y\right) \left({y}^{2} - 2 x\right)}{{\left({y}^{2} - 2 x\right)}^{3}}$

This simplifies to

$\frac{{d}^{2} y}{\mathrm{dx}} ^ \left\{2\right\} = \frac{12 {x}^{2} {y}^{2} - 16 x y - 2 {x}^{4} y - 2 x {y}^{4}}{{\left({y}^{2} - 2 x\right)}^{3}}$

This equation would allow you to find the second derivative of this curve at any point $\left(x , y\right)$ on the curve (such as $\left(x , y\right) = \left(0 , - 1\right)$) where you are not dividing by zero in this last equation.

Here's a picture of the curve defined by the original equation ${x}^{3} + {y}^{3} = 6 x y - 1$. One very interesting thing about it is that it is in two pieces!