Why Does the Partial Derivative of a Sum Cancel Out?

[Ryan187]

Why the summation of the following function will be canceled out when we take the partial derivative with respect to the x_i?
Notice that x_i is the sub of (i), which is the same lower limit of the summation! Can someone, please explain in details?

 

[PeroK]

The $x_i$ are, I assume, independent variables. If $i \ne j$ then $\frac{\partial x_j}{\partial x_i} = 0$.

Note that the answer should be $2x_i$.

 

[Ryan187]

Thank you PeroK! Even if x_i is an independent variable. why the summation is canceled out? can you explain the derivative in detail, please? and you're right the answer is 2x_i not x_i

 

[PeroK]

Thank you PeroK! Even if x_i is an independent variable. why the summation is canceled out? can you explain the derivative in detail, please? and you're right the answer is 2x_i not x_i

The summation is not needed because most terms in the sum are zero.

 

[PeroK]

PS it might help is you use a different dummy variable for the definition of $y$. Use $j = 1$ to $n$ instead of $i$, for example.

 

[etotheipi]

@Ryan187 if you replace the free index $i$ with a concrete number, can you for instance evaluate$$\frac{\partial}{\partial x_3} \sum_{k=1}^5 ({x_k}^2)= \frac{\partial}{\partial x_3} ({x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2 + {x_5}^2)$$assuming the $x_i$ are independent variables?

 

[fresh_42]

explain

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

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

 

[Mark44]

@Ryan187 if you replace the free index $i$ with a concrete number, can you for instance evaluate$$\frac{\partial}{\partial x_3} \sum_{k=1}^5 ({x_k}^2)= \frac{\partial}{\partial x_3} ({x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2 + {x_5}^2)$$assuming the $x_i$ are independent variables?

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

 

[Ryan187]

So, in this case, I am kind of differentiating over one term that's why the sum will be canceled out?

 

[fresh_42]

So, in this case, I am kind of dif\displaystyle{}ferentiating over one term that's why the sum will be canceled out?

No. You differentiate all terms. $\displaystyle{\dfrac{\partial}{\partial x_i}\left(\sum_{j=1}^N x_j^2\right) = \sum_{j=1}^N \dfrac{\partial}{\partial x_i} (x_j^2)}$

 

[etotheipi]

I'll use my example again$$\begin{align*}\frac{\partial}{\partial x_3} \sum_{k=1}^5 ({x_k}^2)&= \frac{\partial}{\partial x_3} ({x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2 + {x_5}^2)\\ \\&= \frac{\partial}{\partial x_3}({x_1}^2) + \frac{\partial}{\partial x_3}({x_2}^2) + \frac{\partial}{\partial x_3}({x_3}^2)+ \frac{\partial}{\partial x_3}({x_4}^2)+\frac{\partial}{\partial x_3}({x_5}^2) \\ \\

&= 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

 

[fresh_42]

I see I've been beaten to it by @fresh_42

I have these shortcuts on the keyboard:

\displaystyle{}
\dfrac{}{}
\left(\right)
\partial

 

[Ryan187]

but in this case, we derive with respect to x3 (i=3) which is one term, but the original derivative is with respect to (x_i) which is the whole i's values?!

 

[fresh_42]

but in this case, we derive with respect to x3 (i=3) which is one term, but the original derivative is with respect to (x_i) which is the whole i's values?!

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.

 

[Ryan187]

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