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Easy differentiation?

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    Im trying to differentiate

    [v-\sum_{i=1}^{n}q_{i}-c_{1}]q_{1}-F

    With respect to q1.

    I know that the answer is:

     v-(\sum_{i=1}^{n}q_{i})-c_{1}+q_{1}(-1)

    But I dont know where the final term comes from (the -q1)

    Could please help, I'm pretty sure this is really simple but ive been staring at it for ages and just dont know where the -q1 comes from.
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    (Original post by RAFecon)
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    You're using, or should be, the product rule on the bracket term in your original expression. The derivative of everything in the brackets is simply -1.
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    Expand the summation term by term for the 1st three terms, say. Then work out the derivatives. You'll see where the -q1 come from then.
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    (Original post by steve10)
    Expand the summation term by term for the 1st three terms, say. Then work out the derivatives. You'll see where the -q1 come from then.
    Yes I can see where it comes from now if I expand the summation, thanks. But I would like to know automatically when using the product rule.

    (Original post by ghostwalker)
    You're using, or should be, the product rule on the bracket term in your original expression. The derivative of everything in the brackets is simply -1.
    I think I'm missing something. How do the brackets differentiate to -1 when you differentiate w.r.t. q1
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    (Original post by RAFecon)
    I think I'm missing something. How do the brackets differentiate to -1 when you differentiate w.r.t. q1
    Differentiating what's inside the brackets term by term. There's only one term in q1, and that's the -q1 under the summation sign (giving -1).

    It's the same as steve10 is suggesting, but doing it in your head.
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    (Original post by ghostwalker)
    Differentiating what's inside the brackets term by term. There's only one term in q1, and that's the -q1 under the summation sign (giving -1).

    It's the same as steve10 is suggesting, but doing it in your head.
    oh, ok. thanks

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