# Lost in second derivative simplification

My book has the following problem:

Find the second derivative of y for $x^2 + xy + y^2 = 3$

$y'' = -\dfrac{ 6(x^2 + xy + y^2) }{ x + 2y }$

But then how do they get here?

$y'' = -\dfrac{ 18 }{ (x + 2y)^3}$

Updated

Assuming the $x^2 + xy + y^2$ is substituted in, then shouldn't it read

$y'' = -\dfrac{ 18 } { x + 2y } = -(\dfrac{ 3 }{ (x + 2y)})^3$

or something like that? I don't see how the bottom gets cubed but the top stays 18

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Note that in the numerator you have $x^2 + xy + y^2$, which is nothing but $3$.

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...Thank you! It seemed like magic...though now I don't see how the denominator gets to be cubed all of a sudden, but I will think about it some more. –  Henry Nov 2 '12 at 3:08

I must have a different method than your book since I never reached the same step as you, but here is my derivation if it helps:

$x^2 + xy + y^2 = 3$

Find $\dfrac{dy}{dx}$:

$2x + [(x)(y') + (y)(1)] + 2y(y') = 0$

$2x + xy' + y + 2yy' = 0$

$(2x + y) + y'(x + 2y) = 0$

Solve for $y'$:

$y' = -\dfrac{ 2x + y }{ x + 2y }$

Find $y''$:

$y'' = -\dfrac{ (x + 2y)(2 + y') - (2x + y)(1 + 2y') }{ (x + 2y)^2 }$

$y'' = -\dfrac{ (2x + xy' + 4y + 2yy') - (2x + 4xy' + y + 2yy') }{ (x + 2y)^2 }$

$y'' = -\dfrac{ 2x + xy' + 4y + 2yy' - 2x - 4xy' - y - 2yy' }{ (x + 2y)^2 }$

$y'' = -\dfrac{ -3xy' + 3y }{ (x + 2y)^2 }$

Substitute for $y'$:

$y'' = -\dfrac{ -3x(-\dfrac{ 2x + y }{ x + 2y }) + 3y }{ (x + 2y)^2 }$

$y'' = -\dfrac{ \dfrac{ 6x^2 + 3xy }{ x + 2y } + 3y }{ (x + 2y)^2 }$

$y'' = -\dfrac{ \dfrac{ 6x^2 + 3xy + 3y(x + 2y) }{ x + 2y } }{ (x + 2y)^2 }$

$y'' = -(\dfrac{ 6x^2 + 3xy + 3xy + 6y^2 }{ x + 2y })(\dfrac{ 1 }{ (x + 2y)^2 })$

$y'' = -\dfrac{ 6x^2 + 6xy + 6y^2 }{ (x + 2y)^3 }$

$y'' = -\dfrac{ 6(x^2 + xy + y^2) }{ (x + 2y)^3 }$

As other users noted, replace $x^2 + xy + y^2 = 3$ from the original equation:

$y'' = -\dfrac{ 6(3) }{ (x + 2y)^3 }$

$y'' = -\dfrac{ 18 }{ (x + 2y)^3 }$

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Substituting $x^2+xy+y^2=3$ into your result.

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