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Math Help - for some reason I cant figure out this derivative :

  1. #1
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    for some reason I cant figure out this derivative :

    heres the problem : d/dx (4x^4 -2x)/4x^4 + 2x)

    and heres my work, let me know if you see something wrong. a * means multiply.

    = 4x^4-2x * (4x^4+2x)^-1
    = 16x^3-2 * (4x^4+2x)^-1 + 16x^3+2 * -(4x^4+2x)^-2 * 4x^4-2x
    =(16x^3-2) / (4x^4+2x) + (16x^3+2) / (-4x^4-2x)
    = factored out a 2 from the tops and a 2x from the bottom
    =8x^3-1 / 2x^4+x + 8x^3+1 / 2x^4-x.
    = added the 2 terms together (by painfully multiplying the bottoms together ect.)
    = 2x^5+2x^4+x^2 / -4x^8-4x^10-x^2
    = factor out a x^2
    = 2x^3+2x^2+1 / -4x^6-4x^8 - 1.

    However im like 99% sure im wrong. What am I doing wrong??
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  2. #2
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    Quote Originally Posted by frankinaround View Post
    heres the problem : d/dx (4x^4 -2x)/4x^4 + 2x)

    and heres my work, let me know if you see something wrong. a * means multiply.

    = 4x^4-2x * (4x^4+2x)^-1
    = 16x^3-2 * (4x^4+2x)^-1 + 16x^3+2 * -(4x^4+2x)^-2 * 4x^4-2x
    =(16x^3-2) / (4x^4+2x) + (16x^3+2) / (-4x^4-2x)
    = factored out a 2 from the tops and a 2x from the bottom
    =8x^3-1 / 2x^4+x + 8x^3+1 / 2x^4-x.
    = added the 2 terms together (by painfully multiplying the bottoms together ect.)
    = 2x^5+2x^4+x^2 / -4x^8-4x^10-x^2
    = factor out a x^2
    = 2x^3+2x^2+1 / -4x^6-4x^8 - 1.

    However im like 99% sure im wrong. What am I doing wrong??
    try simplifying first ...

    \dfrac{4x^4 - 2x}{4x^4 + 2x} = \dfrac{2x(2x^3 - 1)}{2x(2x^3 + 1)} = \dfrac{2x^3 - 1}{2x^3 + 1}

    now ... why are you fooling with the product rule when the quotient rule allows you to avoid keeping track of those nasty negative exponents?

    \dfrac{d}{dx} \left(\dfrac{2x^3 - 1}{2x^3 + 1}\right) = \dfrac{(2x^3+1)(6x^2) - (2x^3-1)(6x^2)}{(2x^3+1)^2} = \dfrac{6x^2[(2x^3+1) - (2x^3-1)]}{(2x^3+1)^2} = \dfrac{12x^2}{(2x^3+1)^2}
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  3. #3
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    because i just convert it to product rule because that way I dont have to memorize a million rules and i figure its the same thing. Thanks because I see thats the right answer.

    Anyway, any idea as to what I did wrong? I spent a long time doing it over and over and it drove me nuts.
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  4. #4
    Super Member TheChaz's Avatar
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    Quote Originally Posted by frankinaround View Post
    heres the problem : d/dx (4x^4 -2x)/4x^4 + 2x)

    and heres my work, let me know if you see something wrong. a * means multiply.

    = 4x^4-2x * (4x^4+2x)^-1
    = 16x^3-2 * (4x^4+2x)^-1 + 16x^3+2 * -(4x^4+2x)^-2 * 4x^4-2x
    =(16x^3-2) / (4x^4+2x) + (16x^3+2) / (-4x^4-2x)
    = factored out a 2 from the tops and a 2x from the bottom
    =8x^3-1 / 2x^4+x + 8x^3+1 / 2x^4-x.
    = added the 2 terms together (by painfully multiplying the bottoms together ect.)
    = 2x^5+2x^4+x^2 / -4x^8-4x^10-x^2
    = factor out a x^2
    = 2x^3+2x^2+1 / -4x^6-4x^8 - 1.

    However im like 99% sure im wrong. What am I doing wrong??
    In red. Some funky algebra. Looks like you canceled a "...+2x" factor with a "... -2x" factor...
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