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Thread: Chain Rule

  1. #1
    Junior Member
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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
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    Hello, Becky!

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


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

    I proceeded this way:

    $\displaystyle f'(x)\:=\3x+1)^3(2)(1-2x)$(-2) $\displaystyle - (1-2x)^2(3)(3x+1)^2(3)$
    . . . . . . . . . . all over denominator-squared

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

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


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

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

    . . . . . . .$\displaystyle 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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