# How do you find (d^2y)/(dx^2) given x^3+y^3=18xy?

Jul 28, 2016

${y}^{' '} \left(x\right) = \frac{2 x y \left(x\right) \left(216 + {x}^{3} - 18 x y \left(x\right) + y {\left(x\right)}^{3}\right)}{6 x - y {\left(x\right)}^{2}} ^ 3$

#### Explanation:

Calling

$f \left(x , y\right) = {x}^{3} + {y}^{3} \left(x\right) - 18 x y \left(x\right) = 0$

(df)/(dx)=3 x^2 - 18 y(x) - 18 x y'(x) + 3 y(x)^2 y'(x) = 0

giving

$y ' \left(x\right) = \frac{{x}^{2} - 6 y \left(x\right)}{6 x - y {\left(x\right)}^{2}}$

Now

d/(dx)((df)/(dx))=6 x - 36 y'(x) + 6 y(x) y'(x)^2 - 18 x y^{''}(x)+ 3 y(x)^2 y^{''}(x)=0

Solving for ${y}^{' '} \left(x\right)$ we have

y^('')(x) = (2 (x - 6 y'(x)+ y(x) y'(x)^2))/( 6 x - y(x)^2)

and after substituting $y ' \left(x\right)$ we obtain finally

${y}^{' '} \left(x\right) = \frac{2 x y \left(x\right) \left(216 + {x}^{3} - 18 x y \left(x\right) + y {\left(x\right)}^{3}\right)}{6 x - y {\left(x\right)}^{2}} ^ 3$

Jul 28, 2016

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 x y \left({y}^{3} + 216 - 18 x y + {x}^{3}\right)}{6 x - {y}^{2}} ^ 3$

#### Explanation:

We first find the first derivative, paying careful attention to $y \left(x\right)$ where we need to apply the chain rule:

$\frac{d}{\mathrm{dx}} \left({x}^{3} + {y}^{3} = 18 x y\right)$

$\rightarrow 3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 18 y + 18 x \frac{\mathrm{dy}}{\mathrm{dx}}$

Rearrange to obtain expression for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

(dy)/(dx) = (3x^2 - 18y)/(18x - 3y^2)" " color(red)("Equation 1")

Simplify:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} - 6 y}{6 x - {y}^{2}}$

Differentiate again using quotient rule:

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

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{\textcolor{b l u e}{12 {x}^{2}} - 2 x {y}^{2} - 36 x \frac{\mathrm{dy}}{\mathrm{dx}} + \textcolor{\mathmr{and} a n \ge}{6 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}} - \textcolor{b l u e}{6 {x}^{2}} + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} + 36 y - \textcolor{\mathmr{and} a n \ge}{12 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}}}{6 x - {y}^{2}} ^ 2$

Some stuff cancels which is nice.

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{6 {x}^{2} - 2 x {y}^{2} - 36 x \frac{\mathrm{dy}}{\mathrm{dx}} - 6 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} + 36 y}{6 x - {y}^{2}} ^ 2$

Now we want to sub in equation 1:

$= \frac{6 {x}^{2} - 2 x {y}^{2} - 36 x \left(\frac{{x}^{2} - 6 y}{6 x - {y}^{2}}\right) - 6 {y}^{2} \left(\frac{{x}^{2} - 6 y}{6 x - {y}^{2}}\right) + 2 {x}^{2} y \left(\frac{{x}^{2} - 6 y}{6 x - {y}^{2}}\right) + 36 y}{6 x - {y}^{2}} ^ 2$

Make every term in the numerator over a common denominator:

$= \frac{6 {x}^{2} \frac{6 x - {y}^{2}}{6 x - {y}^{2}} - 2 x {y}^{2} \frac{6 x - {y}^{2}}{6 x - {y}^{2}} - 36 x \left(\frac{{x}^{2} - 6 y}{6 x - {y}^{2}}\right) - 6 {y}^{2} \left(\frac{{x}^{2} - 6 y}{6 x - {y}^{2}}\right) + 2 {x}^{2} y \left(\frac{{x}^{2} - 6 y}{6 x - {y}^{2}}\right) + 36 y \frac{6 x - {y}^{2}}{6 x - {y}^{2}}}{6 x - {y}^{2}} ^ 2$

Can move this down to the full denominator:

$= \frac{6 {x}^{2} \left(6 x - {y}^{2}\right) - 2 x {y}^{2} \left(6 x - {y}^{2}\right) - 36 x \left({x}^{2} - 6 y\right) - 6 {y}^{2} \left({x}^{2} - 6 y\right) + 2 {x}^{2} y \left({x}^{2} - 6 y\right) + 36 y \left(6 x - {y}^{2}\right)}{6 x - {y}^{2}} ^ 3$

Expanding:

$= \frac{\textcolor{b l u e}{36 {x}^{3}} - 6 {x}^{2} {y}^{2} - 12 {x}^{2} {y}^{2} + 2 x {y}^{4} - \textcolor{b l u e}{36 {x}^{3}} + 216 x y - 6 {x}^{2} {y}^{2} + \textcolor{\mathmr{and} a n \ge}{36 {y}^{3}} + 2 {x}^{4} y - 12 {x}^{2} {y}^{2} + 216 x y - \textcolor{\mathmr{and} a n \ge}{36 {y}^{3}}}{6 x - {y}^{2}} ^ 3$

Again, we can cancel a few things, marked in colour.

We now collect like terms to obtain:

$= \frac{2 x {y}^{4} + 432 x y - 36 {x}^{2} {y}^{2} + 2 {x}^{4} y}{6 x - {y}^{2}} ^ 3$

Factor to finally obtain that:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = \frac{2 x y \left({y}^{3} + 216 - 18 x y + {x}^{3}\right)}{6 x - {y}^{2}} ^ 3$