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Math Help - A single quick differentation

  1. #1
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    A single quick differentation

    Answer at the back of the book is different than what I come out. Im pretty sure there are different ways to get it, but If you would differentiate...
    F(x)=3x^2 (5-x)^4


    how would you get it to be f '(x)=(21x^2 - 14x)(x^3 -2x +1)^6 ?

    tyvm!
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by 3deltat View Post
    Answer at the back of the book is different than what I come out. Im pretty sure there are different ways to get it, but If you would differentiate...
    F(x)=3x^2 (5-x)^4


    how would you get it to be f '(x)=(21x^2 - 14x)(x^3 -2x +1)^6 ?

    tyvm!
    That solution is not possible. I went ahead and did some of the work (before noticing that the solution is not possible) so I'll post it anyways. However, the original function was a 6th degree polynomial (if you have multiplied it all out), but the solution you posted is a 20th degree polynomial. That's clearly wrong.

    f(x)=3x^2(5-x)^4

    f'(x)=\frac{d}{dx}[3x^2]*(5-x)^4+3x^2\frac{d}{dx}[(5-x)^4]

    f'(x)=6x(5-x)^4+3x^2[4(5-x)^3*(-1)]

    f'(x)=6x(5-x)^4-12x^2(5-x)^3

    Now, we can factor 6x(5-x)^3

    f'(x)=6x(5-x)^3[(5-x)-2x]

    exc...
    Last edited by ecMathGeek; May 28th 2007 at 05:25 PM. Reason: Thanks Jhevon
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    That solution is not possible. I went ahead and did some of the work (before noticing that the solution is not possible) so I'll post it anyways. However, the original function was a 6th degree polynomial (if you have multiplied it all out), but the solution you posted is a 20th degree polynomial. That's clearly wrong.

    f(x)=3x^2(5-x)^4

    f'(x)=\frac{d}{dx}[3x^2]*(5-x)^4+3x^2\frac{d}{dx}[(5-x)^4]

    f'(x)=6x(5-x)^4+3x^2[4(5-x)^3*(-1)]

    f'(x)=6x(5-x)^4-12x^4(5-x)^3

    Now, we can factor 6x(5-x)^3

    f'(x)=6x(5-x)^3[(5-x)-2x^3]

    exc...
    the 12x^4 should be 12x^2 but that doesn't detract from what you are saying.

    when we take the derivative of a polynomial, we cut the power down by 1
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