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Math Help - Find derivative using limit definition

  1. #1
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    Find derivative using limit definition

    Find the derivative of f(z)=z^4-z^2 using the following definition:

    f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}

    I tried

    f'(z_0)=\lim_{z\to z_0}\frac{(z^4-z^2)-(z_0^4-z_0^2)}{z-z_0}

    =\lim_{z\to z_0}\frac{z^2(z+1)(z-1)-z_0^2(z_0+1)(z_0-1)}{z-z_0}

    but am unclear how to proceed.
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  2. #2
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    Re: Find derivative using limit definition

    Ragnarok

    Substitute z with z=z0+δz and do all the necessary expansions and simplifications .then the denominator will become δz .
    factorize the numerator and cancel δz . then the limit will show you the final result that must be 4z0^3-2z0
    try it
    Last edited by MINOANMAN; February 28th 2013 at 07:58 AM.
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  3. #3
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    Re: Find derivative using limit definition

    Thanks Minoanman, but I was specifically instructed to use the above "alternative" definition of derivative as opposed to the one using f(z_0+\Delta z)-f(z_0). Could anyone help me with the algebra/tell me what I'm doing wrong?
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  4. #4
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    Re: Find derivative using limit definition

    Quote Originally Posted by Ragnarok View Post
    Find the derivative of f(z)=z^4-z^2 using the following definition:

    f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}

    I tried

    f'(z_0)=\lim_{z\to z_0}\frac{(z^4-z^2)-(z_0^4-z_0^2)}{z-z_0}

    =\lim_{z\to z_0}\frac{z^2(z+1)(z-1)-z_0^2(z_0+1)(z_0-1)}{z-z_0}

    but am unclear how to proceed.

    \frac{(z^4-z^2)-(z_0^4-z_0^2)}{z-z_0}=\frac{(z^4-z_0^4)-(z^2-z_0^2)}{z-z_0}

    Now factor. The idea is divide off the z-z_0 factor.
    Thanks from Ragnarok
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  5. #5
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    Re: Find derivative using limit definition

    Brilliant, thank you!
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  6. #6
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    Re: Find derivative using limit definition

    Hello, Ragnarok!

    You started correctly . . .


    \text{Find the derivative of }f(x)\,=\,x^4-x^2\,\text{ at }x=a

    \text{ using this definition: }\:f'(a)\:=\:\lim_{x\to a}\frac{f(x)-f(a)}{x-a}

    I like to break it up into three steps.

    [1]\;f(x) - f(a) \;=\; (x^4 - x^2) - (a^4-a^2)

    . . . . . . . . . . . =\; (x^4-a^4) - (x^2 - a^2)

    . . . . . . . . . . . =\;(x-a)(x+a)(x^2+a^2) - (x-a)(x+a)

    . . . . . . . . . . . =\;(x-a)\,\big[(x+a)(x^2+a^2) - (x+a)\big]


    [2]\;\frac{f(x)-f(a)}{x-a} \;=\;\frac{(x-a)\big[(x+a)(x^2+a^2) - (x+a)\big]}{x-a}

    . . . . . . . . . . . =\;(x+a)(x^2+a^2) - (x+a)


    [3]\;\lim_{x\to a}\frac{f(x)-f(a)}{x-a} \;=\;\lim_{x\to a}\big[(x+a)(x^2+a^2) - (x+a)\big]

    . . . . . . . . . . . . . . =\; (a+a)(a^2+a^2) - (a+a)

    . . . . . . . . . . . . . . =\;(2a)(2a^2) - 2a


    Therefore: . . f'(a) \;=\;4a^3 - 2a
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