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Math Help - Chain Rule

  1. #1
    Junior Member
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    Mar 2006
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    Post Chain Rule

    We are supposed to differentiate the given function and simplify the answer. I'm not sure if I did this correctly.

    F(x) = (1-2x)^2/(3x+1)^3

    I proceeded this way:
    (3x+1)^3(2)(1-2x) - (1-2x)^2(3)(3x+1)^2(3)=

    (3x+1)^2(1-2x)[2(3x+1)-9(1-2x)] =

    (3x+1)^2(1-2x)[(6x+2)-(9-18x)]

    (3x+1)^2(1-2x)(24x-7)
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  2. #2
    Super Member

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    Hello, Becky!

    Sorry, you made two omissions in the first step . . .


    f(x) \:= \:\frac{(1-2x)^2}{(3x+1)^3}

    I proceeded this way:

    3x+1)^3(2)(1-2x)" alt="f'(x)\:=\3x+1)^3(2)(1-2x)" />(-2) - (1-2x)^2(3)(3x+1)^2(3)
    . . . . . . . . . . all over denominator-squared

    It should have looked like this:
    . . f'(x)\:=\:\frac{(3x+1)^3(2)(1-2x)(-2) - (1-2x)^2(3)(3x+1)^2(3)}{(3x+1)^6}

    . . . . . . = \;\frac{-4(3x+1)^3(1-2x) - 9(1-2x)^2(3x+1)^2}{(5x+1)^6}


    Factor: . f'(x)\:=\:\frac{-(3x+1)^2(1-2x)\,\left[4(3x+1) + 9(1-2x)\right]}{(3x+1)^6}

    . . . . . . . f'(x) \:=\:\frac{-(1-2x)(12x + 4 + 9 - 18x)}{(3x+1)^4}

    . . . . . . . f'(x)\:=\:\boxed{\frac{-(1-2x)(13 - 6x)}{(3x+1)^4}}\;\text{ or }\;\boxed{\frac{(2x-1)(13-6x)}{(3x+1)^4}}


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