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Math Help - chain rule

  1. #1
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    chain rule

    find f '(x)

    f(x)=4/(3xto2nd-2x+1)to3rd
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by homerb View Post
    find f '(x)

    f(x)=4/(3xto2nd-2x+1)to3rd
    Note that \frac{4}{(3x^2-2x+1)^3}=4(3x^2-2x+1)^{-3}

    Do you know how to apply the chain rule?

    \frac{\,d}{\,dx}\left[f(g(x))\right]=f'(g(x))\cdot g'(x)

    Let f(x)=4x^{-3} and let g(x)=3x^2-2x+1

    Can you take it from here?

    --Chris
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  3. #3
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    no I can't
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    \frac{\,d}{\,dx}\left[f(g(x))\right]=f'(g(x))\cdot g'(x)

    Let f(x)=4x^{-3} and let g(x)=3x^2-2x+1

    Then we see that f'(x)=-12x^{-4} and g'(x)=6x-2

    So when we apply the chain rule, we see that \frac{\,d}{\,dx}\left[4(3x^2-2x+1)^{-3}\right]=-12(3x^2-2x+1)^{-4}\cdot (6x-2)

    Can you take it from here and simplify this?

    --Chris
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  5. #5
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    I try to do it, but I can't. This is my first time doing this. If you can simplify this more
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by homerb View Post
    I try to do it, but I can't. This is my first time doing this. If you can simplify this more
    The farthest it can be simplified:

    -12(3x^2-2x+1)^{-4}\cdot (6x-2)=-\frac{24(3x-1)}{(3x^2-2x+1)^4}

    --Chris
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  7. #7
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    homer you times the -12 by the 6x-2 for the numerator and the equation in the parenthesis goes in the denominator and it becomes a positive exponent because you put it in the denominator
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