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Math Help - Differentiation

  1. #1
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    Differentiation

    This is a topic i am really struggling to get my head around.

    1. the expression (x+a)^-1 can be approximated for small values of a by the relationship

    (x+a)^-1 = (1/x) - (a/x^2) + (a^2/x^3)

    Using the relationship, derive an expression for the rate of change of the function f(x)=1/x by the 'first principles' differentiation method.
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  2. #2
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    Quote Originally Posted by daniel123 View Post
    This is a topic i am really struggling to get my head around.

    1. the expression (x+a)^-1 can be approximated for small values of a by the relationship

    (x+a)^-1 = (1/x) - (a/x^2) + (a^2/x^3)

    Using the relationship, derive an expression for the rate of change of the function f(x)=1/x by the 'first principles' differentiation method.
    f'(x)=lim \frac{f(x+a)-f(x)}{(x+a)-x}, where a-> 0

    now substitute f(x+a) by (x+a)^{-1} = \frac{1}{x} - \frac{a}{x^2} + \frac{a^2}{x^3} and f(x) by 1/x in the f'(x) formula.
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  3. #3
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    Ok thankyou, i have an answer, which is,

    =-1/x^2
    Last edited by mr fantastic; October 3rd 2010 at 12:25 PM. Reason: Moved questions to new thread, added / so that the fraction was obvious.
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