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    Hi

    Could someone help me with the following equation please?

    Please see attached.

    The attached image also contains the answer to the problem. I just don't understand how the equation is rearranged to make Q the subject.

    If anyone could help would be much appreciated.

    Thanks
    Michael
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    They did not make Q the subject. They differentiated and then solved dTc/dQ=0.

    First you might like to write TC=HC+RC+\frac{1}{2}QC_h+DC_0 Q^{-1}.

    Now differentiate to find \frac{dTC}{dQ}
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    Babymaths tank you for your response. Maths really isn't my strongest subject. Could you explain how to differentiate \frac{dTC}{dQ} please?

    Thanks
    Michael
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    (Original post by mstephens88)
    Babymaths tank you for your response. Maths really isn't my strongest subject. Could you explain how to differentiate \frac{dTC}{dQ} please?

    Thanks
    Michael
    OK.

    If I gave you \displaystyle y=a+bx+cx^{-1} (where a,b and c are constants) and asked you to find \displaystyle  \frac{dy}{dx} could you do that?
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    No sorry, I don't really understand the principle behind deriving an equation
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    (Original post by mstephens88)
    No sorry, I don't really understand the principle behind deriving an equation
    What course are you doing and what maths have you covered before?

    It appears from the question you've posted that your course expects you to know some basic calculus, e.g. differentiation!
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    I'm doing product design, but this particular module is engineering management. We did do a little maths in the first year but that was 4 or so years ago.
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    (Original post by mstephens88)
    I'm doing product design, but this particular module is engineering management. We did do a little maths in the first year but that was 4 or so years ago.
    Basic differentiation:
    \ y = ax^n + bx^m + c

\Rightarrow \dfrac{dy}{dx} = nax^{n-1} + mbx^{m-1}
    So if you swap y for TC and x for Q . . .
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    Ok thanks, I understand that but I just don't understand how to apply that to the original equation.
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    I always remember it as: multiply by the power, then take one from the power
    And if you need to know integration: add one to the power, divide by the new power i have maths today as well :/ but C1 and C2 good luck!


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    (Original post by mstephens88)
    Ok thanks, I understand that but I just don't understand how to apply that to the original equation.
    So, in this case, \ Q^n = nQ^{n-1} Where n is any number.
 
 
 
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