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Thread: Calculate : f'(0)

  1. #1
    Newbie SADANE ANTAR's Avatar
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    Calculate : f'(0)

    f is the function :
    $\displaystyle
    f(x+y)=f(x)+f(y)+xy^{2}+x^{2}y
    $
    and : $\displaystyle {f}'(0)=1$
    Calculate :$\displaystyle {f}'(4)$
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by SADANE ANTAR View Post
    f is the function :
    $\displaystyle
    f(x+y)=f(x)+f(y)+xy^{2}+x^{2}y
    $
    and : $\displaystyle {f}'(0)=1$
    Calculate :$\displaystyle {f}'(4)$
    If the derivative exists at $\displaystyle x$ then:

    $\displaystyle
    f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x-h)}{2h}=$ $\displaystyle \lim_{h \to 0}\frac{f(x)+f(h)+xh^2+x^2h-f(x)-f(-h)+xh^2+x^2h}{2h}=$ $\displaystyle
    \lim_{h \to 0}\frac{f(h)-f(-h)+2x^2h}{2h}=f'(0)+x^2$

    CB
    Last edited by CaptainBlack; Dec 14th 2009 at 06:46 AM.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    If the derivative exists at $\displaystyle x$ then:

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

    CB
    According to ...

    Derivative -- from Wolfram MathWorld

    ... the 'more symmetrical' notation of the derivative is...

    $\displaystyle f^{'} (x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$

    ... in any case congratulations for Your elegant procedure! ...



    Merry Christmas from Italy

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by chisigma View Post
    According to ...

    Derivative -- from Wolfram MathWorld

    ... the 'more symmetrical' notation of the derivative is...

    $\displaystyle f^{'} (x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x-h)}{2h}$

    ... in any case congratulations for Your elegant procedure! ...



    Merry Christmas from Italy

    $\displaystyle \chi$ $\displaystyle \sigma$

    Oppss ... ( a hangover from a earlier attempt with a non-symmetric form) It should be correct now

    CB
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  5. #5
    MHF Contributor chisigma's Avatar
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    If we accept the 'more symmetrical' notation of the derivative...

    $\displaystyle f^{'} (x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x-h)}{2h}$ (1)

    ...do we also have to accept that for a function like $\displaystyle f(x)=|x|$ is $\displaystyle f^{'} (0)=0$?... or not? ...



    Merry Christmas from Italy

    $\displaystyle \chi$ $\displaystyle \sigma$
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by chisigma View Post
    If we accept the 'more symmetrical' notation of the derivative...

    $\displaystyle f^{'} (x) = \lim_{h \rightarrow 0} \frac{f(x+h) - f(x-h)}{2h}$ (1)

    ...do we also have to accept that for a function like $\displaystyle f(x)=|x|$ is $\displaystyle f^{'} (0)=0$?... or not? ...



    Merry Christmas from Italy

    $\displaystyle \chi$ $\displaystyle \sigma$
    $\displaystyle f'(0)=1$ implies that the derivative exists at $\displaystyle 0$, which excludes the posibility of something like $\displaystyle |x|$. Also this can be done with the derivative from the right being $\displaystyle 1$ at $\displaystyle x=0$ (as implied in my comment about an earlier version using an asymmetric form for the difference quotient).

    CB
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