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Math Help - Simple fraction derivative.

  1. #1
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    Simple fraction derivative.

    Hi all,

    I have been asked to differentiate the function f(x)=\frac{5}{x} where (x>0). I have not been asked previously in exercises (or taught for that matter) how to differentiate fractions like this. I have used Mathcad to find the solution -\frac{5}{x^2} hoping it would reveal a method to me, but with no such luck. I would really appreciate any help on this. I think it may be a rule for powers prior to finding the derivative that I am missing.

    Regards,

    Shayne.
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  2. #2
    Behold, the power of SARDINES!
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    It is just the general power rule for derivatives but you need to use some algebra first.

    f(x)=\frac{5}{x}=5x^{-1} \implies f'(x)=-5x^{-2}=-\frac{5}{x^2}

    You could also use the quotient rule but that would be way too much work.
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  3. #3
    Super Member TheChaz's Avatar
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    Quote Originally Posted by Hydralisk View Post
    Hi all,

    I have been asked to differentiate the function f(x)=\frac{5}{x} where (x>0). I have not been asked previously in exercises (or taught for that matter) how to differentiate fractions like this. I have used Mathcad to find the solution -\frac{5}{x^2} hoping it would reveal a method to me, but with no such luck. I would really appreciate any help on this. I think it may be a rule for powers prior to finding the derivative that I am missing.

    Regards,

    Shayne.
    f(x)=\frac{5}{x} = 5x^{-1}

    By the power rule, the derivative of this is

    (-1)*5x^{-1 -1} = -5x^{-2} = - \frac{5}{x^2}
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  4. #4
    Super Member TheChaz's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    It is just the general power rule for derivatives but you need to use some algebra first.

    f(x)=\frac{5}{x}=5x^{-1} \implies f'(x)=-5x^{-2}=-\frac{5}{x^2}
    Yeah, what he said!
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  5. #5
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    Thanks for that,

    I very much appreciate it.

    Shayne.
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  6. #6
    A Plied Mathematician
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    I'm not sure I agree that the quotient rule is SO much more work. I often find that it's a bit more calculus up front, but can somtimes be less algebra later. My working:

    \left(\frac{5}{x}\right)'=\frac{(x)(0)-(5)(1)}{(x)^{2}}=-\frac{5}{x^{2}}.

    Where I especially see a savings in effort is when taking derivatives of functions like this:

    \frac{\sin(x)}{\sqrt{1-x^{2}}}.

    If you flip the denominator upstairs and use the product rule, you'll have to turn right around, often, and get a common denominator. Using the quotient rule saves you that step.

    My two cents.
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