How Do You Differentiate a Double Summation?

A double summation is still a sum: it adds terms indexed by two indices. If the limits are fixed and the sum is finite, differentiation is straightforward—you differentiate each term. Infinite double sums need an extra check, because differentiating term by term can fail without suitable convergence conditions.

Finite double sums

Suppose

$$F(x)=\sum_{i=m}^{n}\sum_{j=p}^{q} f_{ij}(x),$$

where the indices and their limits do not depend on $x$. Since this contains only finitely many terms, the ordinary sum rule for derivatives gives

$$F'(x)=\sum_{i=m}^{n}\sum_{j=p}^{q} f'_{ij}(x).$$

You do not differentiate the indices, and you do not change the limits. This works because differentiating a finite sum is the same as adding the derivatives of its terms.

The rule also applies when one or both index ranges are written with a fixed finite set rather than consecutive integers. The essential point is that there are finitely many summands and the indexing set stays fixed as $x$ varies.

Worked example

Differentiate

$$F(x)=\sum_{i=1}^{2}\sum_{j=1}^{3}x^{i+j}.$$

For each pair $(i,j)$, the derivative of $x^{i+j}$ is $(i+j)x^{i+j-1}$. Therefore

$$F'(x)=\sum_{i=1}^{2}\sum_{j=1}^{3}(i+j)x^{i+j-1}.$$

To check the result, expand the original sum:

$$F(x)=x^2+x^3+x^4+x^3+x^4+x^5 =x^2+2x^3+2x^4+x^5.$$

Differentiating this polynomial gives

$$F'(x)=2x+6x^2+8x^3+5x^4.$$

For instance, $F'(1)=21$. Expanding is not necessary in general; it is a useful way to verify that the indices and exponents were handled correctly.

Infinite double sums

For an infinite double series, writing down the term-by-term derivative is only a candidate. You must justify exchanging differentiation with the infinite summation. A standard sufficient condition is:

• each summand $f_{ij}$ is continuously differentiable on an interval;
• the derivative series $\sum_i\sum_j f'_{ij}(x)$ converges uniformly there, using a consistent interpretation of the double sum; and
• the original series converges at at least one point in the interval.

Under these conditions, the sum defines a differentiable function and its derivative is the sum of the derivatives. A convenient way to check uniform convergence is the Weierstrass M-test: find constants $M_{ij}$ such that $|f'_{ij}(x)|\le M_{ij}$ throughout the interval and $\sum_i\sum_j M_{ij}$ is finite. These are sufficient conditions, not the only possible ones.

Infinite-series example

For $|x|<1$, consider

$$G(x)=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}x^{i+j}.$$

The two geometric series give

$$G(x)=\left(\sum_{i=0}^{\infty}x^i\right) \left(\sum_{j=0}^{\infty}x^j\right) =\frac{1}{(1-x)^2}.$$

Thus $G'(x)=2/(1-x)^3$. Termwise differentiation gives the same result:

$$G'(x)=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}(i+j)x^{i+j-1}.$$

To justify this locally, work on any closed subinterval $|x|\le r$ with $0<r<1$. The absolute values of the derivative terms are bounded by terms whose total sum is

$$\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}(i+j)r^{i+j-1} =\frac{2}{(1-r)^3},$$

which is finite. The M-test therefore gives uniform convergence of the derivative series on that subinterval. Since every point with $|x|<1$ lies in such a subinterval, termwise differentiation is valid throughout the stated domain.

What if the limits depend on x?

A sum such as $\sum_{i=1}^{N}\sum_{j=1}^{M}f_{ij}(x)$ still has fixed limits if $N$ and $M$ are fixed integers, so the finite-sum rule applies. If a limit is an integer-valued function of $x$, such as $N=\lfloor x\rfloor$, the number of terms can jump as $x$ changes. The resulting function may be discontinuous at those jumps, so ordinary term-by-term differentiation does not automatically apply. Analyze the intervals where the bounds stay constant and check the jump points separately.

This differs from differentiating an integral with variable endpoints: sums do not acquire the boundary terms from the Leibniz rule for integrals.

Common mistakes

• Assuming an infinite sum behaves like a finite one. Convergence of the original series alone does not guarantee that you can differentiate it term by term.
• Differentiating the indices. With fixed limits, indices are labels, not functions of $x$; differentiate the summands.
• Changing the summation bounds without cause. Differentiation does not remove or add terms when the index ranges are fixed.
• Ignoring the domain. A power series or other infinite sum may converge, and allow termwise differentiation, only on a particular interval or region.