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Math Help - Finding a f'(x) in a fraction

  1. #1
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    Finding a f'(x) in a fraction

    I don't know how I should go about solving down on this problem for f'(x):

    f(x)=\frac{-6x^3+3x^2-2x+1}{x}

    If someone could show me in a step-by-step fashion that would be great.
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  2. #2
    Super Member Random Variable's Avatar
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    You could use the quotient rule or you could just divide each term in the numerator by x and then take the derivative.
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  3. #3
    Senior Member Spec's Avatar
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    Quotient rule - Wikipedia, the free encyclopedia

    In this case, it's easier to just split it up.

    f(x)=\frac{-6x^3+3x^2-2x+1}{x}=-6x^2+3x-2+\frac{1}{x}
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  4. #4
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    I'm still a bit confused on the quotient rule, can you go a bit further with the math explaination?

    Thanks a lot!
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  5. #5
    Super Member Random Variable's Avatar
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    Quote Originally Posted by hemi View Post
    I'm still a bit confused on the quotient rule, can you go a bit further with the math explaination?

    Thanks a lot!
    For this problem there is no reason to use the quotient rule. But here's what it would look like:

    if  f(x) = \frac {g(x)}{h(x)} then  f'(x) = \frac {h(x)g'(x)-g(x)h'(x)}{[h(x)]^{2}}

    so  f'(x) = \frac {x \frac {d}{dx} (-6x^{3}+3x^{2}-2x+1) - (-6x^3+3x^{2}-2x+1) \frac {d}{dx}x} {x^{2}}

     = \frac {x(-18x^{2}+6x-2) - (-6x^{3}+3x^{2}-2x+1)}{x^{2}}

     = \frac {-12x^{3} +3x^{2}-1}{x^{2}}

     = -12x + 3 - \frac {1}{x^{2}}


    The quotient rule is just an extension of the product rule.
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