Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to, (45) and (46) (p. 6),

differention of $$\alpha = \sum_{j=1}^n\sum_{i=1}^n a_{ij} x_i x_j $$

with respect to the k-th element of x yields:

$$\frac{\partial\alpha}{\partial x_k} = \sum_{j=1}^n a_{kj} x_j + \sum_{i=1}^n a_{ik} x_i $$

Note that a does not depend on x.

How is this result obtained?

From differentiation with summation symbol, I understood how to derive one summation. The function above seems to be of form f(g(x)) to me, so I would apply the chain rule. But how can the result contain a + then, indicating some form of the product rule was used?

share|cite|improve this question
up vote 1 down vote accepted

Much simpler: $$ \frac{\partial}{\partial x_k} \sum_{i, j} a_{i j} x_i x_j = \sum_{i, j} a_{i j} \left( \frac{\partial x_i}{\partial x_k} x_j + x_i \frac{\partial x_j}{\partial x_k} \right) = \sum_j a_{k j} x_j + \sum_i a_{i k} x_i $$

share|cite|improve this answer

Note that all the terms $a_{ij}x_i x_j$ will vanish when neither $i$ nor $j$ is equal to $k$. So all the remaining terms are of the form $a_{ik}x_i x_k$ and $a_{kj}x_k x_j$, which when differentiated with respect to $x_k$ will yield respectively $a_{ik}x_i$ and $a_{kj} x_j$, as in the announced formula.

share|cite|improve this answer
What about when $j=k$? – gt6989b Mar 28 '13 at 21:09
$j=k$ corresponds to the case $a_{ik}x_i x_k$ – Glougloubarbaki Mar 28 '13 at 21:11
I mean $i=j$ -- then differentiation will have a different outcome since you are differentiating $a_{kk}x_k^2$. – gt6989b Mar 28 '13 at 21:15
then just put $i=k$ in the first kind of terms, or $j=k$ in the second. the factor 2 will appear because you'll get one term of each kind – Glougloubarbaki Mar 28 '13 at 22:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.