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# 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:

δyx = δux + δvx

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,..., fn. Then

d/dx (∑i=1 to n fi(x)) = d/dx [f1(x) + f2(x) + ... + fn(x)] = d/dx(f1(x)) + d/dx(f2(x)) + ... + d/dx(fn(x))
so
d/dx (∑i=1 to n fi(x)) = ∑i=1 to n (d/dx(fi(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(fi(x)).

Then

i=1 to k fi(x) = g(x) + fk(x)

and it follows from the proof above that

d/dx (∑i=1 to k fi(x)) = d/dx(g(x)) + d/dx(fk(x))

By the inductive hypothesis,

d/dx(g(x)) = d/dx (∑i=1 to k-1 fi(x)) = ∑i=1 to k-1 d/dx (fi(x))

so

d/dx (∑i=1 to k fi(x)) = ∑i=1 to k-1 d/dx (fi(x)) + d/dx (fk(x)) = ∑i=1 to k d/dx (fi(x))

which ends our proof.

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