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Math Help - Function problem

  1. #1
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    Function problem

    Hey there. I am working on a function problem with which I am having a hard time getting started on.

    Differentiate the following function:

    f(x) = 3x/x^2 + 4

    Once I can get it to polynomial form I'll be good to go, but I'm a little stuck as it stands now. Could somebody give me a nudge to get me started? I think it has something to do with x^-n = 1/x^n.... Thanks.
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  2. #2
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    The derivative of a quotient of functions is given by the following rule:

    \left[\frac{f}{g}\right]'=\frac{f'\cdot g-f\cdot g'}{g^2}.

    I think you can manage from there
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  3. #3
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    I have never seen this rule before.... So my f(x) = 3x/x^2 + 4, but what detremines my g(x)? Thanks again.
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  4. #4
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    Quote Originally Posted by Jools View Post
    I have never seen this rule before.... So my f(x) = 3x/x^2 + 4, but what detremines my g(x)? Thanks again.
    What course are you doing?
    Why were you asked to find the derivative of a quotient if you have never seen the quotient rule?
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  5. #5
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    It's a calculus and vectors pre-university course. We were show the power, constant, sum and difference, and product rules, as well as the chain rule for powers of polynomials...
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  6. #6
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    f(x)=\frac{3x}{x^2+4}

    using te quotient rule

    f`(x) = \frac{3(x^2+4) - 6x^2}{(x^2+4)^2}
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  7. #7
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    If you were given this...

    Why dont you solve it this way then,
     f(x) \frac{3x}{x^2}+4

    You can simply to  f(x) \frac{3}{x} +4

    \frac{3}{x}+4 = 3x^{-1} +4

    And this you probably seen before?
    Or did I miss someting?... is that 4 suppose to be in the denominator? That "hint" you gave in your first post made me think that 4 is not in the term \frac{3x}{x^2}
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  8. #8
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    Quote Originally Posted by Jools View Post
    It's a calculus and vectors pre-university course. We were show the power, constant, sum and difference, and product rules, as well as the chain rule for powers of polynomials...
    \left[ {\frac{f}<br />
{g}} \right]^\prime   = \left[ {fg^{ - 1} } \right]^\prime   = f'g^{ - 1}  + f\left( { - g^{ - 2} g'} \right) = \frac{{f'g - fg'}}<br />
{{g^2 }}
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  9. #9
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    Alright thanks guys. I think I've got it now. I was never shown the quotient rule. My only question is how come the derivative for each of the two last methods is so different?
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  10. #10
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    K... Here's how I did it. It's a little different approach...

    f(x) = 3x / x^2 + 4
    = (3X)(X^2 + 4)^-1

    Then used a combination of the product rule and chain rule to come up with:

    = -6x / (x^2 + 4)^2 + 3/x^2 + 4

    Look right?
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  11. #11
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    You're missing a square. It's -6x^2
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