How to Differentiate a Double Summation Using Dot Notation

A double summation is just a sum taken over two indices. If it has finitely many terms, its index bounds do not depend on the variable, and its summands are differentiable with respect to that variable, you can differentiate it term by term. Dot notation is a shorthand for a time derivative; it does not change the differentiation rule.

The rule for a finite double sum

Suppose each summand is differentiable with respect to time, while the limits $m$ and $n$ are fixed:

$$S(t)=\sum_{i=1}^{m}\sum_{j=1}^{n} f_{ij}(t).$$

Then

$$\dot{S}(t)=\frac{dS}{dt}=\sum_{i=1}^{m}\sum_{j=1}^{n}\frac{df_{ij}}{dt}.$$

The sums contain only finitely many terms, so differentiation and summation can be interchanged. The order of the two sums can also be reversed without changing the result.

A dot over a quantity means its derivative with respect to time: $\dot{S}=dS/dt$. Two dots mean a second time derivative: $\ddot{S}=d^2S/dt^2$. If the variable is not time, ordinary derivative notation such as $dS/dx$ is usually clearer. A dot between vectors, as in $\mathbf{a}\cdot\mathbf{b}$, instead denotes a dot product; context distinguishes the two uses.

Worked example

Let

$$S(t)=\sum_{i=1}^{2}\sum_{j=1}^{3}(it^2+jt).$$

The bounds are fixed, so differentiate the summand while keeping $i$ and $j$ constant:

$$\dot{S}(t)=\sum_{i=1}^{2}\sum_{j=1}^{3}(2it+j).$$

To simplify, there are three copies of $2it$ for each value of $i$, and two copies of the sum $1+2+3$ in the terms involving $j$:

$$\begin{aligned} \dot{S}(t) &=3(2t)(1+2)+2(1+2+3)\\ &=18t+12. \end{aligned}$$

You can check by expanding the original sum first. The terms involving $it^2$ add to $9t^2$, and the terms involving $jt$ add to $12t$, so $S(t)=9t^2+12t$. Differentiating gives the same result, $18t+12$.

When the summand contains products or functions

The rule for sums does not replace the usual differentiation rules. Apply those rules separately to each term. For example, if

$$S(t)=\sum_{i=1}^{m}\sum_{j=1}^{n} a_i(t)b_j(t),$$

then

$$\dot{S}(t)=\sum_{i=1}^{m}\sum_{j=1}^{n}\left(\dot{a}_i(t)b_j(t)+a_i(t)\dot{b}_j(t)\right),$$

by the product rule. If a summand depends on quantities that themselves vary with time, use the chain rule as well. For instance, if $f_{ij}=f_{ij}(x(t),y(t))$, then

$$\frac{df_{ij}}{dt} =\frac{\partial f_{ij}}{\partial x}\dot{x} +\frac{\partial f_{ij}}{\partial y}\dot{y}.$$

What if the bounds vary?

The term-by-term rule above assumes fixed index ranges. A bound such as $\lfloor t\rfloor$ makes the number of terms change at integer values of $t$. The resulting sum may have jumps there, so it need not have an ordinary derivative at those points. You cannot generally apply the Leibniz rule for differentiating an integral to a discrete sum; that is a different operation.

For an infinite double sum, differentiating term by term also requires extra conditions. For example, suitable uniform convergence of the series of derivatives, together with convergence at a point, is a common sufficient condition. Without such conditions, the derivative of the sum need not equal the sum of the derivatives.

Common mistakes

• Differentiating the indices: Treat $i$ and $j$ as constants when they are summation indices. Differentiate the expression that depends on the variable.
• Dropping the product rule: A sum of products still requires the product rule for each product.
• Treating a changing number of terms as fixed: If a bound depends on the variable, first examine where the number of terms changes.
• Reading every dot as a time derivative: A dot above a variable denotes a time derivative in common physics notation; a dot between vectors denotes a dot product.
• Writing a dot over the summation sign without explanation: Prefer $\dot{S}$ for the whole summed quantity, or write the sum of the dotted summands explicitly. This makes clear what is being differentiated.