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Thread: 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 $\displaystyle \frac{4}{(3x^2-2x+1)^3}=4(3x^2-2x+1)^{-3}$

    Do you know how to apply the chain rule?

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

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

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

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

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

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

    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:

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

    --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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