Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My book has the following problem:

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

I follow them this far:

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

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

share|improve this question
add comment

3 Answers

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

share|improve this answer
    
...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
add comment

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:

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

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

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.