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Math Help - implicit differentiation con't

  1. #1
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    implicit differentiation con't

    Problem: x^4(x+y)=y^2(3x-y)

    Solution attempt:

    \frac{d}{dx}[x^4(x+y)]-\frac{d}{dx}[y^2(3x-y)]=0

    [4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3y-y')]=0

    upon distribution of coefficients,

    4x^4+4x^3y+x^4+x^4y'-6xyy'+2y^2y'-3y^2+y^2y'=0

    If I didn't subtract \frac{d}{dx}[y^2(3x-y)] as I did in step 1 I end up with,

    4x^4+4x^3y+x^4+x^4y'=6xyy'+2y^2y'-3y^2+y^2y'

    This is where I am confronted with many mathematically valid yet counter productive steps as I try to solve for y'

    combining like terms I have,

    5x^4+4x^3y+x^4y'=6xyy'+3y^2y'-3y^2

    5x^4+4x^3y+x^4y'=y'(6xy+3y^2)-3y^2

    Answer in book: y'=\frac{3y^2-5x^4-4x^3y}{x^4+3y^2-6xy}

    notes: what gets me is the fact that I have two y' and \frac{y'}{y'}=1
    Last edited by Foxlion; March 10th 2011 at 01:30 PM.
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  2. #2
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    i'm going to combine like termms and see where that gets me.
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  3. #3
    Super Member TheChaz's Avatar
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    Quote Originally Posted by Foxlion View Post
    i'm going to combine like termms and see where that gets me.
    First, you have a sign error.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Foxlion View Post
    Problem: x^4(x+y)=y^2(3x-y)

    Solution attempt:

    \frac{d}{dx}[x^4(x+y)]-\frac{d}{dx}[y^2(3x-y)]=0

    [4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3y-y')]=0
    The last line should read
    [4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3-y')]=0

    -Dan
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  5. #5
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    [4x^3(x+y)+x^4(1+y')] - [2yy'(3x-y)+y^2(3-y')]=0

    -Dan
    is typo, following lines do not take it into account
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  6. #6
    Super Member TheChaz's Avatar
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    I'll just springboard off of Dan's post (and yours...):
    yes, you want to combine like terms. You have, among other things,
    x^4y' - 6xyy' + 2y^2y' +y^2y' = y'(x^4 - 6xy + 3y^2)

    After you move the rest of the terms to the other side and divide by
    (x^4 - 6xy + 3y^2), you'll have the desired result.
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  7. #7
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    <sigh>...I got it...

    x^4(x+y) = y^2(3x-y)
    differentiating both sides...
    4x^3(x+y)+x^4(1+y') = 2yy'(3x-y+y^2(3-y')
    distributing coefficients...
    4x^4+4x^3y+x^4+x^4y' = 6xyy'-2y^2y'+3y^2-y^2y'
    combining like terms...
    5x^4+4x^3y+x^4y' = 6xyy'-3y^2y'+3y^2
    subtracting 3y^2from both sides...
    5x^4+4x^3y+x^4y'-3y^2 = 6xyy'-3y^2y'
    subtracting x^4y' from both sides...
    5x^4+4x^3y-3y^2 = 6xyy'-3y^2y'-x^4y'
    factoring out y' from right...
    5x^4+4x^3y-3y^2 = y'(6xy-3y^2-x^4)
    dividing both sides by  (6xy-3y^2-x^4)...
    \frac{5x^4+4x^3y-3y^2}{6xy-3y^2-x^4} = y'
    now for some rearrangement as well as multiplying both sides by \frac{-1}{-1}...
    y' = \frac{3y^2-5x^4-4x^3y}{x^4+3y^2-6xy}

    Thank you both
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