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Math Help - Differrentiate this function

  1. #1
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    Differrentiate this function

    Hi all, just stuck with a simple equation at the moment and dont really know how to solve it. Hope you guys could assist me! Thanks.

    The function is given as:

    y= 1/C (1 - e^(-C(z - x))

    what is the answer given i have differrentiate: dy/dx

    C is a scalar constant

    Thanks and much appreciated!
    Last edited by Redeemer_Pie; October 15th 2010 at 09:58 AM.
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  2. #2
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    e^(i*pi)'s Avatar
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    Assuming you mean: \dfrac{1}{C}\left(1-e^{-C(z - x)}\right) and that Z and C are constants.

    Expand to give \dfrac{1}{C}\left(1-e^{Cz-Cx}\right) = \dfrac{1}{C} - \dfrac{1}{C}\left(e^{Cz-Cx}\right)

    Using the laws of exponents we can write e^{Cz-Cx} = e^{Cz} \cdot e^{-Cx}

    Group the constants: \dfrac{1}{C} - \dfrac{e^{Cz}}{C}\,e^{-Cx}


    That should be quite easy to differentiate using the chain rule
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  3. #3
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    thanks^

    But im still confused.

    How come the expotential became Cz-Cx not:

    -Cz + Cx?
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  4. #4
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    Can someone give me the final result of the equation?

    Im not too sure whether the constant disappears

    Thanks
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  5. #5
    Senior Member Educated's Avatar
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    As you said before, C is a scalar constant so it can be either positive or negative.

    You won't learn anything if we give you the answer.

    y= \dfrac{\left(1-e^{Cz - Cx}\right)}{C}

    y= \dfrac{1}{C} - \dfrac{e^{Cz - Cx}}{C}

    y= \dfrac{1}{C} - \dfrac{e^{Cz} \cdot e^{-Cx}}{C}

    Can you differentiate from here?
    And no, the constants do not disappear.
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  6. #6
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    i thought the answer will be like:

    1/C - e^Cz . e^-Cx
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  7. #7
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    C is a constant and so is \frac{1}{C}. Its derivative is 0. As Educated told you,
    y= \frac{1}{C}- \frac{1}{C}(e^{Cz}e^{-Cx}

    The derivative of the first \frac{1}{C} is 0.

    What is the derivative of Ae^{-Cx} where A and C are constants?
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