The Sum and Difference Rule

You're on a train doing a steady 30 metres per second, and you stroll towards the front of the carriage at 1.5 metres per second. How fast are you moving over the ground? You don't reach for anything clever — you just add: 30 + 1.5 = 31.5 m/s. The train contributes its rate, your legs contribute theirs, and the rates stack.

The same instinct works everywhere rates live. If your savings account grows by £40 a month and your side business brings in £120 a month, your total wealth grows at £160 a month — each income stream changes at its own rate, and the total changes at the sum of them. Fill a bath with the tap adding 8 litres a minute while the plughole drains 3: the water level changes at 8 - 3 = 5 litres a minute.

A derivative is a rate of change — so this everyday instinct is secretly a theorem of calculus. Differentiation distributes over + and -: the derivative of a sum is the sum of the derivatives, the derivative of a difference is the difference. This page makes that precise, shows exactly why it's true, and — just as important — marks where the shortcut stops working.

The rule

In symbols, the rule says you may differentiate term by term: pull an expression apart at its + and - signs, differentiate each piece with whatever rule fits it, and reassemble with the same signs.

Notice what the rule does not require: f and g don't have to be related to each other in any way. One can be a polynomial and the other a sine wave; each term minds its own business. You never have to wrestle a whole expression at once — break it at the signs, handle the pieces, and the signs carry straight through to the answer.

Why it works: slopes just add

The derivative measures a slope — a rate of change. If at some instant f is climbing at 3 units per step and g is climbing at 2, then their sum f + g climbs at 3 + 2 = 5. Stacking one curve on top of the other adds their heights at every point — and a small step to the right raises each height by its own slope's worth, so the stacked height rises by the two slopes' worth combined. Adding heights everywhere means adding slopes everywhere.

See it live. The faint dashed curves below are f(x) = \sin(x) and the straight line g(x) = cx; the bold curve is their sum. Drag the slider to tilt the line. At c = 0 the bold sum is just the sine wave; make c positive and every part of the sum tilts upward by exactly that much — where the sine was momentarily flat, the sum now has slope c. Push c negative and the whole wave leans downhill instead. The line donates its slope to the sum at every single point:

In symbols, what you're watching is \frac{d}{dx}\big[\sin(x) + cx\big] = \cos(x) + c — the sine's own slope, shifted up or down by the constant c.

Once, honestly: the proof in one line

"Slopes add" is a picture; here is the same fact from the limit definition of the derivative, and it's almost embarrassingly short. Write the difference quotient for the sum f + g, and watch it split at the plus sign:

\frac{\big[f(x+h) + g(x+h)\big] - \big[f(x) + g(x)\big]}{h} = \frac{f(x+h) - f(x)}{h} + \frac{g(x+h) - g(x)}{h}

All we did was regroup the numerator — collect the f-parts together and the g-parts together — and share the h between them. The quotient for the sum is exactly the quotient for f plus the quotient for g, for every step size h. Now let h \to 0: the limit of a sum is the sum of the limits, so the left side becomes (f+g)'(x) while the right becomes f'(x) + g'(x). Done.

Concretely, for f(x) + g(x) = x^2 + x^3 the split looks like this:

\frac{\big[(x+h)^2 + (x+h)^3\big] - \big[x^2 + x^3\big]}{h} = \underbrace{\frac{(x+h)^2 - x^2}{h}}_{\to\ 2x} + \underbrace{\frac{(x+h)^3 - x^3}{h}}_{\to\ 3x^2}

Each brace is a difference quotient you have already tamed on its own — the square's tends to 2x, the cube's to 3x^2 — so

\frac{d}{dx}\big[x^2 + x^3\big] = 2x + 3x^2.

That's the whole justification, and you only ever need to see it once. From here on, "differentiate term by term" is a licence you hold, not a leap of faith — every time you use it, this one-line regrouping is silently doing the work.

A worked example — and its picture

Suppose h(x) = x^2 + 3x. This is a sum of two terms, so we handle each on its own. Using the power rule on x^2 and the constant multiple rule on 3x:

\frac{d}{dx}\big[x^2\big] = 2x, \qquad \frac{d}{dx}\big[3x\big] = 3

Now just add the two results back together:

h'(x) = \frac{d}{dx}\big[x^2 + 3x\big] = 2x + 3

Here is that example drawn out — h(x) = x^2 + 3x (solid) beside its derivative h'(x) = 2x + 3 (dashed). The dashed line crosses zero at x = -1.5, exactly where the solid curve bottoms out and stops falling — the derivative doesn't just come from the curve, it narrates it: negative while the parabola descends, zero at the turn, positive on the climb back up.

A difference is no different. For x^2 - 3x the second term enters with a minus sign and leaves with one: 2x - 3. Same two derivatives, same reassembly — the sign is just along for the ride.

Minding the minus signs

Longer expressions bring more signs, and signs are where marks are lost. The safest habit: read each minus sign as glued to the term that follows it, so the expression is a sum of signed terms. Take

p(x) = x^3 - 5x^2 + 4x - 7

and hear it as "x^3, plus (-5x^2), plus 4x, plus (-7)". Now differentiate the four signed terms one at a time:

\frac{d}{dx}\big[x^3\big] = 3x^2, \quad \frac{d}{dx}\big[-5x^2\big] = -10x, \quad \frac{d}{dx}\big[4x\big] = 4, \quad \frac{d}{dx}\big[-7\big] = 0

The -5x^2 keeps its minus through the whole computation — the constant multiple rule with the constant -5 — and the lone -7 is a constant, so it vanishes entirely. Reassemble:

p'(x) = 3x^2 - 10x + 4.

Note the ending: three terms came out of four, because the constant's derivative is 0 and simply drops off. Shifting a whole curve up or down by 7 changes none of its slopes, so the -7 was never going to leave a fingerprint on the derivative.

A physical check: two cars, one total

Rates adding isn't a formal trick — nature does it constantly. Send two delivery cars out from the depot along the same road. Car A is accelerating from rest, so its distance after t seconds is A(t) = t^2; car B cruises at a steady 3 m/s, so B(t) = 3t. The total distance the fleet has covered is the sum, A(t) + B(t) = t^2 + 3t — the very function from the worked example above.

Now ask how fast that total is growing. Physically it's obvious: each second, car A adds its current speed's worth of metres and car B adds its own, so the total grows at speed of A plus speed of B. Calculus says the same thing term by term:

\frac{d}{dt}\big[t^2 + 3t\big] = \underbrace{2t}_{\text{A's speed}} + \underbrace{3}_{\text{B's speed}}

At t = 4 seconds, car A is doing 8 m/s and car B its usual 3, so the fleet's total distance is growing at 11 m/s. The sum rule is the mathematical shadow of a fact you already believed: independent contributions to a total contribute their rates independently too.

Three traps, in rising order of danger:

The sum rule is your first meeting with one of the biggest ideas in mathematics: linearity. Together with the constant multiple rule it says the derivative is a linear operator — it respects addition and scaling — and linear things are everywhere:

"Break it into pieces, solve the pieces, add" is arguably science's favourite strategy — and it works precisely where the mathematics is linear. When you later meet operations that aren't linear (squaring, multiplying functions, most of real life), you'll know to put the shortcut down.

It chains across many terms

The rule isn't limited to two terms — apply it repeatedly and a long sum comes apart into as many pieces as you like, each differentiated independently while the + and - signs carry straight through:

\frac{d}{dx}\big[f + g - h\big] = f' + g' - h'

This is the engine behind differentiating any polynomial: a polynomial is nothing but powers of x chained together with + and -, so the sum rule reduces it to power-rule snacks. You'll assemble that full recipe next. First, see Sal work through the basic rules: