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Math Help - function --- derivative (help!)

  1. #1
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    function --- derivative (help!)


    but instead of f(x)=x^-2, let f(x) = x^(1/2 )


    So, I did
    [ (x+h)^1/2 - (x)^1/2 ]/ h
    But im not sure how to simplify it. What would be my next step? Thanks alot
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by gomes View Post

    but instead of f(x)=x^-2, let f(x) = x^(1/2 )


    So, I did
    [ (x+h)^1/2 - (x)^1/2 ]/ h
    But im not sure how to simplify it. What would be my next step? Thanks alot
    wait..

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

    now using the definition

    \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}

    \lim_{h \to 0} \frac{{\frac{1}{(x+h)^2}}-{\frac{1}{x^2}}}{h}

    simplify this and get your proof
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  3. #3
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    Quote Originally Posted by gomes View Post

    but instead of f(x)=x^-2, let f(x) = x^(1/2 )
    \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}

    \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}<br />

    \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}<br />

    \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}

    finish it.
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  4. #4
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    Thanks everyone

    Quote Originally Posted by skeeter View Post
    \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}

    \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}<br />

    \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})}<br />

    \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}

    finish it.
    Thanks, I'm having trouble with sin(2x), i think my double angle formulae, i might be doing something wrong/missing something. Any help? Most appreciated!

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  5. #5
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    Sorry, one more, im stuck

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  6. #6
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by gomes View Post
    Sorry, one more, im stuck

    I'm lazy so I wont reproduce this here but look at : http://www.mash.dept.shef.ac.uk/Reso...principles.pdf

    If you have any questions we can answer them here
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  7. #7
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    Quote Originally Posted by AllanCuz View Post
    I'm lazy so I wont reproduce this here but look at : http://www.mash.dept.shef.ac.uk/Reso...principles.pdf

    If you have any questions we can answer them here
    Thanks! i understand it, but what about for sin2x?

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  8. #8
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by gomes View Post
    Thanks! i understand it, but what about for sin2x?

    \lim_{h \to 0} \frac{sin(2x+2h)-sin(2x)}{h}

    = \lim_{h \to 0} \frac{2}{2} \times  \frac{sin(2x+2h)-sin(2x)}{h}

    = \lim_{h \to 0} \frac{2(sin(2x+2h)-sin(2x))}{2h}

    = \lim_{h \to 0} \frac{2(sin(X+H)-sin(X))}{H} where X=2x and H=2h

    = 2cos(X) ; which you proved before doing this.

    = 2cos(2x)
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