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Math Help - Simple Calculus Question

  1. #1
    Newbie evanator's Avatar
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    Simple Calculus Question

    Hi there,

    There is no forum for basic calculus, so I will have to post in the university-level forum. I have been trying to refresh my maths for the last few months (I had forgotten everything, right down to how to do long division.) and I feel as if I have been making good progress.

    I have a question about a very simple differentiation exercise:

    Using the formula

    f'(x) = \displaystyle\lim_{\delta x \to 0} \frac{f(x + \delta x) -f(x)}{\delta x}

    find the derivative of \frac{1}{x^2}

    Here is my attempt at a solution:

    f(x + \delta x) = \frac{1}{(x + \delta x)^2}

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

    = \frac{x^2 - (x^2 + 2x \delta x + (\delta x)^2)}{(x^2 + 2x \delta x + (\delta x)^2)x^2}

    = \frac{\delta x(- \delta x -2x)}{x^4 + 2x^3 \delta x + x^2(\delta x)^2}

    \frac{f(x + \delta x) - f(x)}{\delta x} = \frac{\delta x(- \delta x - 2x)}{x^4 + 2x^3 \delta x + x^2(\delta x)^2} \times \frac{1}{\delta x}

    = \frac{-2x - \delta x}{x^4 + 2x^3 \delta x + x^2(\delta x)^2}

    f'(x) = \displaystyle\lim_{\delta x \to 0} \frac{-2x - \delta x}{x^4 + 2x^3 \delta x + x^2(\delta x)^2}

    = \frac{-2x}{x^4}

    = \frac{-2}{x^3}

    It would be great if someone would tell me where I am going wrong.

    Regards,

    Evanator
    Last edited by evanator; July 7th 2010 at 05:09 AM.
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  2. #2
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    Quote Originally Posted by evanator View Post
    Hi there,

    There is no forum for basic calculus, so I will have to post in the university-level forum. I have been trying to refresh my maths for the last few months (I had forgotten everything, right down to how to do long division.) and I feel as if I have been making good progress.

    I have a question about a very simple differentiation exercise:

    Using the formula

    f'(x) = \displaystyle\lim_{\delta x \to 0} \frac{f(x + \delta x) -f(x)}{\delta x}

    find the derivative of \frac{1}{x^2}

    Here is my attempt at a solution:

    f(x + \delta x) = \frac{1}{(x + \delta x)^2}

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

    ..............
    you forgot x in second step ,
    f(x + \delta x) - f(x) = \frac{1}{x^2 + 2x \delta x + (\delta x)^2}  - \frac{1}{x^2}

    now continue....
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  3. #3
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    mr fantastic's Avatar
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    Quote Originally Posted by evanator View Post
    Hi there,

    There is no forum for basic calculus, so I will have to post in the university-level forum.

    [snip]
    Well done. Thankyou. Someone with some basic common sense. And you even showed your work.
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  4. #4
    Newbie evanator's Avatar
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    I amended the first post. It all comes out now. Thanks for the help, ramiee2010. It is much appreciated.
    What are you referring to, mr fantastic?
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  5. #5
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    Quote Originally Posted by evanator View Post
    I amended the first post. It all comes out now. Thanks for the help, ramiee2010. It is much appreciated.
    What are you referring to, mr fantastic?
    He's saying a lot of people post calculus questions in the pre-calculus forum just because they happen to be in high school and not university.

    And he's also giving you praise for showing your working, as a lot of people also just post a question and beg for others to do the work for them.
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