Sum rule in differentiation

The sum rule in differentiation is possibly the most useful rule in differentiation. The sum rule in integration follows from it. The rule itself is a direct consequence of differentiation from first principles.

The sum rule tells us that for two functions u and v:

d/dx(u + v) = du/dx + dv/dx

This rule also applies to subtraction and to additions and subtractions of more than two functions (d/dx(u + v + w + ...) = du/dx + dv/dx + dw/dx + ...). Furthermore, it applies to summations.

Proof

Let y be a function given by the sum of two functions u and v, such that:

y = u + v

Now let y, u and v be increased by small increases δy, δu and δv respectively. Hence:

y + δy = (u + δu) + (v + δv) = u + v + δu + δv = y + δu + δv

So:

δy = δu + δv

Now divide throughout by δx:

δy/δx = δu/δx + δv/δx

Let δx tend to 0:

dy/dx = du/dx + dv/dx

Now recall that y = u + v, giving the sum rule in differentiation:

d/dx(u + v) = du/dx + dv/dx

The rule can be extended to subtraction, as follows:

d/dx(u - v) = d/dx(u + (-v)) = du/dx + d/dx(-v)

Now use the special case of the constant factor rule in differentiation with k=-1 to obtain:

d/dx(u - v) = du/dx + (-dv/dx) = du/dx - dv/dx

Therefore, the sum rule can be extended so it "accepts" addition and subtraction as follows:

d/dx(u ± v) = du/dx ± dv/dx

The sum rule in differentiation can be used as part of the derivation for both the sum rule in integration and linearity of differentiation.

Generalization to sums

Assume we have some set of functions f₁, f₂,..., f_n. Then

d/dx (∑_{i=1 to n} f_i(x)) = d/dx [f₁(x) + f₂(x) + ... + f_n(x)] = d/dx(f₁(x)) + d/dx(f₂(x)) + ... + d/dx(f_n(x))

so

d/dx (∑_{i=1 to n} f_i(x)) = ∑_{i=1 to n} (d/dx(f_i(x))

In other words, the derivative of any sum of functions is the sum of the derivatives of those functions.

This follows easily by induction; we have just proven this to be true for n = 2. Assume it is true for all n < k, then define

g(x) = ∑_{i=1 to k-1}(f_i(x)).

Then

∑_{i=1 to k} f_i(x) = g(x) + f_k(x)

and it follows from the proof above that

d/dx (∑_{i=1 to k} f_i(x)) = d/dx(g(x)) + d/dx(f_k(x))

By the inductive hypothesis,

d/dx(g(x)) = d/dx (∑_{i=1 to k-1} f_i(x)) = ∑_{i=1 to k-1} d/dx (f_i(x))

so

d/dx (∑_{i=1 to k} f_i(x)) = ∑_{i=1 to k-1} d/dx (f_i(x)) + d/dx (f_k(x)) = ∑_{i=1 to k} d/dx (f_i(x))

which ends our proof.