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Math Help - derivative proof

  1. #1
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    derivative proof

    Hi, can anyone help me on this problem?

    Suppose that instead of the usual definition of the derivative Df(x) , we define a new kind of derivative, D*f(x), by the forumla

    D*f(x) = lim h---> 0 {f^2 (x+h) - f^2 ( x)}/h where f^2 (x) = [f(x)]^2

    A) Derive formulas for computing the derivative D* of a sum, difference, product, and quotient. (only one solution is needed, from that I think I can figure out the rest)

    B) Express D*f(x) in terms of Df(x)


    Thanks
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by storchfire1X View Post
    Hi, can anyone help me on this problem?

    Suppose that instead of the usual definition of the derivative Df(x) , we define a new kind of derivative, D*f(x), by the forumla

    D^*f(x) = lim h---> 0 {f^2 (x+h) - f^2 ( x)}/h where f^2 (x) = [f(x)]^2

    A) Derive formulas for computing the derivative D* of a sum, difference, product, and quotient. (only one solution is needed, from that I think I can figure out the rest)
    Here's my attempt at (A), I will only do the sum part.

    Consider the differentiable functions f(x) \mbox { and } g(x)

    D^*[f(x) + g(x)] = \lim_{h \to 0} \frac {[f(x + h) + g(x + h)]^2 - [f(x) + g(x)]^2}{h} .....now i was thinking about using the difference of two squares here, but i couldn't really see the fruits of that labor, maybe someone could do it that way and make it work out to something nicer.

    = \lim_{h \to 0} \frac {f^2(x + h) + 2f(x + h)g(x + h) + g^2 (x + h) - f^2(x) - 2f(x)g(x) - g^2 (x)}{h}

    = \lim_{h \to 0} \frac {[f^2(x + h) - f^2 (x) ]+ [g^2 (x + h) - g^2 (x)] + 2[f(x + h)g(x + h) - f(x)g(x)]}{h}

    = \lim_{h \to 0} \frac {f^2(x + h) - f^2(x)}{h} + \lim_{h \to 0} \frac {g^2 (x + h) - g^2 (x)}{h} + 2 \lim_{h \to 0} \frac {f(x + h)g(x + h) - f(x)g(x)}{h}

    = D^*f(x) + D^*g(x) + 2 \frac {d}{dx}f(x)g(x)





    B) Express D^*f(x) in terms of Df(x)


    Thanks
    D^*f = \lim_{h \to 0} \frac {f^2(x + h) - f^2(x)}{h}

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

    = \lim_{h \to 0}[f(x + h) + f(x)] \frac {d}{dx}  f(x)
    Last edited by Jhevon; June 7th 2007 at 02:42 PM.
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    Consider the differentiable functions f(x) \mbox { and } g(x)
    D^*f = \lim_{h \to 0} \frac {f^2(x + h) - f^2(x)}{h}

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

    = \lim_{h \to 0}[f(x + h) + f(x)] \frac {d}{dx}  f(x)
    Note that if f is differentiable then then it is continuous.
    Thus = \lim_{h \to 0}[f(x + h) + f(x)]=2f(x)
    and = \lim_{h \to 0}[f(x + h) + f(x)] \frac {d}{dx}  f(x) = 2f(x) \frac {d}{dx} f(x)
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Plato View Post
    Note that if f is differentiable then then it is continuous.
    Thus = \lim_{h \to 0}[f(x + h) + f(x)]=2f(x)
    and = \lim_{h \to 0}[f(x + h) + f(x)] \frac {d}{dx}  f(x) = 2f(x) \frac {d}{dx} f(x)
    do you have any suggestions for the first one Plato?
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  5. #5
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    Hello, storchfire1X!

    Here's the Product Rule . . .


    Suppose that instead of the usual definition of the derivative Df(x) ,
    we define a new kind of derivative, D^*\!f(x), by the forumla:

    D^*\!f(x) \;= \;\lim_{h\to0}\frac{f^2(x+h) - f^2(x)}{h} . where f^2(x) \,= \,[f(x)]^2
    Let P(x) \:=\:f(x)\!\cdot\!g(x)

    The numerator of the difference quotient is:
    . . [f(x+h)\!\cdot\!g(x+h)]^2 - [f(x)\!\cdot\!g(x)]^2 \;=\;f^2(x+h)\!\cdot\!g^2(x+h) - f^2(x)\!\cdot\!g(x)


    Subtract and add f^2(x+h)\!\cdot\!g^2(x):
    . . f^2(x+h)\!\cdot\!g^2(x+h) - f^2(x+h)\!\cdot\!g^2(x) - f^2(x)\!\cdot\!g^2(x) + f^2(x+h)\!\cdot\!g^2(x)


    Factor:
    . . f^2(x+h)\cdot[g^2(x+h) - g^2(x)] \;+ \;g^2(x)\cdot[f^2(x+h) - f^2(x)]


    Divide by h:
    . . f^2(x+h)\!\cdot\!\frac{g^2(x+h) - g^2(x)}{h} \;+\;g^2(x)\!\cdot\!\frac{f^2(x+h) - f^2(x)}{h}


    Take the limit:
    . . . . \lim_{h\to0}\left[f^2(x+h)\!\cdot\!\frac{g^2(x+h) - g^2(x)}{h} \;+\;g^2(x)\!\cdot\!\frac{f^2(x+h) - f^2(x)}{h}\right]

    . . = \;\lim_{h\to0}\left[f^2(x+h)\right]\cdot\underbrace{\lim_{h\to0}\left[\frac{g^2(x+h) - g^2(x)}{h}\right]}_{\text{This is }D^*g(x)} \;+\;g^2(x)\cdot\underbrace{\lim_{h\to0}\left[\frac{f^2(x+h) - f^2(x)}{h}\right]}_{\text{This is }D^*f(x)}


    Therefore: . D^*\!P(x) \;=\;f^2(x)\cdot D^*\!g(x) \:+\:D^*\!f(x)\cdot g(x)

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  6. #6
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    Quote Originally Posted by Soroban View Post
    Therefore: ^*\!f(x)\cdot g(x) " alt="D^*\!P(x) \;=\;f^2(x)\cdot D^*\!g(x) \:+\^*\!f(x)\cdot g(x) " />
    I am sure that this is a typo.
    ^*\!f(x)\cdot g^2(x) " alt="D^*\!P(x) \;=\;f^2(x)\cdot D^*\!g(x) \:+\^*\!f(x)\cdot g^2(x) " />
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