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Notice that x_i is the sub of (i), which is the same lower limit of the summation! Can someone, please explain in details?

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- Thread starter Ryan187
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No. ## i ## is just one specific index, an arbitrary one, but a specific one. That's why we have to distinguish this index from the summation index which runs over all. One of them matches ##j=i## whereas all others do not: ##j\neq i##. The former yields ##2x_i##, the latter yields the zeros.Got it now! Thanks, guys, you're amazing!f

- #1

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Notice that x_i is the sub of (i), which is the same lower limit of the summation! Can someone, please explain in details?

- #2

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Note that the answer should be ##2x_i##.

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- #5

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Differentiation is a linear function: ##\dfrac{\partial}{\partial x_j}(\alpha \cdot f(x)+\beta\cdot g(x))=\alpha\cdot \dfrac{\partial}{\partial x_j}f(x)+\beta\cdot \dfrac{\partial}{\partial x_j} g(x)##.explain

Here we only have a longer sum: ##x_1^2+x_2^2+\ldots+x_N^2##

- #8

Mentor

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This is the key -- expanding the sum, and then differentiating with respect to one of the terms in the sum.

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- #10

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No. You differentiate all terms. ##\displaystyle{\dfrac{\partial}{\partial x_i}\left(\sum_{j=1}^N x_j^2\right) = \sum_{j=1}^N \dfrac{\partial}{\partial x_i} (x_j^2)}##So, in this case, I am kind of dif\displaystyle{}ferentiating over one term that's why the sum will be canceled out?

- #11

&= 0 + 0 + \frac{\partial}{\partial x_3}({x_3}^2) + 0 + 0\end{align*}$$I see I've been beaten to it by @fresh_42

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- #13

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- #14

- 18,505

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No. ## i ## is just one specific index, an arbitrary one, but a specific one. That's why we have to distinguish this index from the summation index which runs over all. One of them matches ##j=i## whereas all others do not: ##j\neq i##. The former yields ##2x_i##, the latter yields the zeros.

- #15

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Got it now! Thanks, guys, you're amazing! It is my first thread to this community!

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