Sum and Difference Rule for Derivatives and Integrals

What Is The Sum And Difference Rule?

The sum and difference rule states that the derivative of a sum of two differentiable functions is the sum of their derivatives, and the derivative of a difference is the difference of their derivatives. In words: you may differentiate term by term, and the plus and minus signs stay where they are.

If $f$ and $g$ are both differentiable at $x$, then

$$\frac{d}{dx}\left[f(x) + g(x)\right] = \frac{d}{dx}f(x) + \frac{d}{dx}g(x)$$

$$\frac{d}{dx}\left[f(x) - g(x)\right] = \frac{d}{dx}f(x) - \frac{d}{dx}g(x)$$

Here $f$ and $g$ are functions of the input $x$, and $\frac{d}{dx}$ means "the rate of change with respect to $x$". The condition matters: each piece must have a derivative at the point. When both do, the sum and the difference automatically have derivatives there too.

The rule runs in two directions you will use constantly. Forward, it breaks a long expression into short pieces. In reverse, it lets you integrate term by term, which the integral section below makes precise. It also extends to any finite number of terms:

$$\frac{d}{dx}\left[f_1(x) + f_2(x) + \cdots + f_n(x)\right] = \frac{d}{dx}f_1(x) + \frac{d}{dx}f_2(x) + \cdots + \frac{d}{dx}f_n(x)$$

That extension follows by applying the two-term rule repeatedly (a short induction), so a polynomial with ten terms needs nothing new.

Why Does The Derivative Of A Sum Equal The Sum Of The Derivatives?

"Why does the derivative of a sum equal the sum of the derivatives?" is a question readers ask on Quora, and the answer comes straight out of the limit definition of the derivative. The difference quotient of a sum splits into two difference quotients, and limits respect addition.

Let $s(x) = f(x) + g(x)$. By definition,

$$\frac{d}{dx}s(x) = \lim_{h \to 0} \frac{s(x+h) - s(x)}{h}$$

$$= \lim_{h \to 0} \frac{\left[f(x+h) + g(x+h)\right] - \left[f(x) + g(x)\right]}{h}$$

Regroup the four terms so each function's change sits together:

$$= \lim_{h \to 0} \left[\frac{f(x+h) - f(x)}{h} + \frac{g(x+h) - g(x)}{h}\right]$$

Both limits on the right exist, because $f$ and $g$ are differentiable at $x$. The sum law, one of the basic limit laws, says the limit of a sum is the sum of the limits when both limits exist:

$$= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} + \lim_{h \to 0} \frac{g(x+h) - g(x)}{h}$$

$$= \frac{d}{dx}f(x) + \frac{d}{dx}g(x)$$

The difference rule has the same proof with a minus sign carried through every line. A second route is shorter once you know that a constant factor passes through a derivative: write $f - g = f + (-1),g$, apply the sum rule, then pull out the $-1$.

So the rule is not a convention someone chose. It is the limit sum law wearing different clothes, which is also why the same pattern shows up for limits, derivatives and integrals alike.

What Does The Sum And Difference Rule Mean Geometrically?

Geometrically, the rule says slopes add. Stack the graph of $g$ on top of the graph of $f$, point by point, and the steepness of the new graph at any $x$ is the steepness of $f$ there plus the steepness of $g$ there.

Take $f(x) = x^2$ and $g(x) = x$. At $x = 1$, the tangent to $y = x^2$ has slope $2$ and the line $y = x$ has slope $1$. The graph of $y = x^2 + x$ at $x = 1$ has tangent slope $2 + 1 = 3$, which you can confirm by differentiating $x^2 + x$ to get $2x + 1$ and substituting $x = 1$.

The rates picture is just as useful. Two pipes fill a tank, one at a rate that varies with time and one at a steady rate; the water level rises at the sum of the two rates. Open a drain and the level changes at the inflow rate minus the outflow rate. Any quantity built by adding or subtracting other quantities changes at the matching sum or difference of their rates.

How Does The Sum And Difference Rule Work With The Constant Multiple Rule?

The two rules combine into one statement called linearity of differentiation. A Brainly question asks students to "write down the following derivative rules: (a) the sum and difference rules (b) the constant multiple" rule, and together they read:

$$\frac{d}{dx}\left[a,f(x) + b,g(x)\right] = a,\frac{d}{dx}f(x) + b,\frac{d}{dx}g(x)$$

where $a$ and $b$ are constants and $f$ and $g$ are differentiable at $x$. Setting $a = b = 1$ gives the sum rule; $a = 1$ and $b = -1$ gives the difference rule; $b = 0$ gives the constant multiple rule on its own.

Linearity is what makes polynomials easy. Every one of the polynomial functions is a sum of constant multiples of powers of $x$, so you split it into terms, apply the power rule to each power, and keep the coefficients and signs:

$$\frac{d}{dx}\left(4x^3 - 2x^2 + 9\right) = 4 \cdot 3x^2 - 2 \cdot 2x + 0 = 12x^2 - 4x$$

The same pattern covers mixed functions. Sines, exponentials and logarithms each have their own derivative, and the sum and difference rule lets you handle them side by side without any interaction between the terms.

Does The Sum And Difference Rule Work For Products And Quotients?

No. The derivative of a product is not the product of the derivatives, and the derivative of a quotient is not the quotient of the derivatives. Quora readers ask "What is the rule of differentiation of a product of two functions?" exactly because the sum pattern looks like it should carry over, and it does not.

One example settles it. Take $x \cdot x$:

$$\frac{d}{dx},(x \cdot x) = \frac{d}{dx}\left(x^2\right) = 2x$$

$$\frac{d}{dx},(x) \cdot \frac{d}{dx},(x) = 1 \cdot 1 = 1$$

The two answers disagree at every $x$ except $x = \frac{1}{2}$, so "multiply the derivatives" is wrong. Quotients fail the same way: $\frac{x^2}{x} = x$ (for $x \neq 0$) has derivative $1$, while dividing the derivatives gives $\frac{2x}{1} = 2x$.

Products need the product rule, and quotients need the quotient rule. The table below is a quick way to choose:

There is often a way back to the sum rule. Expand a product first, $(x + 1)^2 = x^2 + 2x + 1$, or split a fraction whose denominator is a single term, $\frac{x^2 + 3}{x} = x + \frac{3}{x}$. After rewriting, the expression is a sum again and term-by-term differentiation works. Splitting only works over a single-term denominator, though: $\frac{x}{x + 1}$ cannot be split into $\frac{x}{x} + \frac{x}{1}$.

Does The Sum Rule Need Both Functions To Be Differentiable?

The hypothesis guarantees the conclusion, but it is not required for the sum to be differentiable. If $f$ and $g$ are both differentiable at $x$, then $f + g$ is too. The reverse can fail.

Take $f(x) = |x|$ and $g(x) = -|x|$. Neither is differentiable at $x = 0$, where each graph has a sharp corner (see absolute value functions). Their sum is $f(x) + g(x) = 0$, a constant, and its derivative at $0$ is $0$. So you cannot read the rule backwards and split the derivative of a sum into parts that do not exist; the corners cancel only after adding.

This direction tripped me up the first time too; the rule is a one-way guarantee, and the differentiability of a function has to be checked piece by piece before the split is allowed. For infinite sums the situation is harder still, which is why term-by-term differentiation of a power series comes with its own convergence conditions.

What Is The Sum And Difference Rule For Integrals?

Integration inherits the rule in reverse: the integral of a sum is the sum of the integrals, and the integral of a difference is the difference of the integrals. For indefinite integrals,

$$\int \left[f(x) \pm g(x)\right] dx = \int f(x),dx \pm \int g(x),dx$$

and the result carries one combined constant $+,C$. The reason is the derivative rule itself: if $F$ is an antiderivative of $f$ and $G$ is an antiderivative of $g$, then the derivative of $F \pm G$ is $f \pm g$, so $F \pm G + C$ is the antiderivative of $f \pm g$.

$$\int \left(4x^3 - 6x + 5\right) dx = x^4 - 3x^2 + 5x + C$$

Check by differentiating: $\frac{d}{dx}\left(x^4 - 3x^2 + 5x + C\right) = 4x^3 - 6x + 5$.

For definite integrals the same split is one of the standard properties of definite integrals, and its picture is stacked area. The area under $y = x^2 + x$ from $0$ to $2$ is the area under $y = x^2$ plus the area under $y = x$ over the same interval:

$$\int_{0}^{2} \left(x^2 + x\right) dx = \int_{0}^{2} x^2,dx + \int_{0}^{2} x,dx = \frac{8}{3} + 2 = \frac{14}{3}$$

The integral version has the same blind spot as the derivative version. The integral of a product is not the product of the integrals: $\int x \cdot x,dx = \frac{x^3}{3} + C$, while $\left(\frac{x^2}{2}\right)\left(\frac{x^2}{2}\right) = \frac{x^4}{4}$.

Examples Of The Sum And Difference Rule

Example 1

Find the derivative of $y = x^3 + x^2$, and its value at $x = 2$.

Differentiate each term and keep the plus sign.

$$\frac{dy}{dx} = \frac{d}{dx}\left(x^3\right) + \frac{d}{dx}\left(x^2\right)$$

$$= 3x^2 + 2x$$

At $x = 2$:

$$\frac{dy}{dx} = 3(4) + 2(2) = 16$$

Final answer: $\frac{dy}{dx} = 3x^2 + 2x$, which equals $16$ at $x = 2$.

Example 2

Differentiate $f(x) = 7x^5 + 3x^2 - 4x + 12$.

Split into four terms, use the constant multiple rule and the power rule on each, and keep every sign.

$$\frac{d}{dx}f(x) = \frac{d}{dx}\left(7x^5\right) + \frac{d}{dx}\left(3x^2\right) - \frac{d}{dx},(4x) + \frac{d}{dx},(12)$$

$$= 35x^4 + 6x - 4 + 0$$

$$= 35x^4 + 6x - 4$$

Final answer: $35x^4 + 6x - 4$.

Example 3

Differentiate $y = (x^2 + 1)(x - 3)$.

The sum rule works so smoothly that the natural next move is to treat the product the same way and multiply the two derivatives:

$$\frac{d}{dx}\left(x^2 + 1\right) \cdot \frac{d}{dx},(x - 3) = 2x \cdot 1 = 2x$$

Test that at $x = 0$. The tempting answer gives slope $0$. But $y(0) = (1)(-3) = -3$ and $y(0.001) \approx -2.999$, so the graph rises by about $0.001$ over a run of $0.001$, a slope close to $1$, not $0$. The shortcut has failed, because the sum rule says nothing about products.

The rescue is to rewrite the product as a sum first. Expand:

$$y = x^3 - 3x^2 + x - 3$$

Now the sum and difference rule applies term by term:

$$\frac{dy}{dx} = 3x^2 - 6x + 1 - 0$$

$$= 3x^2 - 6x + 1$$

At $x = 0$ this gives $1$, which matches the numerical check. The product rule gives the same result without expanding.

Final answer: $\frac{dy}{dx} = 3x^2 - 6x + 1$.

Example 4

Differentiate $g(x) = \sin x - e^x + \ln x$ for $x > 0$.

Each term has a standard derivative. The differentiation of trigonometric functions gives $\cos x$ for $\sin x$; the derivatives of exponential functions give $e^x$ for $e^x$; and the derivatives of logarithmic functions give $\frac{1}{x}$ for $\ln x$.

$$\frac{d}{dx}g(x) = \frac{d}{dx},(\sin x) - \frac{d}{dx}\left(e^x\right) + \frac{d}{dx},(\ln x)$$

$$= \cos x - e^x + \frac{1}{x}$$

Final answer: $\cos x - e^x + \frac{1}{x}$, valid for $x > 0$.

Example 5

Find the equation of the tangent line to $y = \frac{x^2 + 3}{x}$ at $x = 1$.

Rewrite the single-term denominator as a sum, then differentiate term by term.

$$y = x + 3x^{-1}$$

$$\frac{dy}{dx} = 1 - 3x^{-2} = 1 - \frac{3}{x^2}$$

At $x = 1$, the point is $y = \frac{1 + 3}{1} = 4$ and the slope is $1 - 3 = -2$. Using point-slope form, as in tangent line equations:

$$y - 4 = -2,(x - 1)$$

$$y = -2x + 6$$

Final answer: the tangent line is $y = -2x + 6$.

Example 6

A workshop's revenue is $R(q) = 40q - 0.01q^2$ dollars and its cost is $C(q) = 2000 + 12q$ dollars for $q$ items. Find the rate of change of profit at $q = 1000$ and at $q = 1500$, and the output where it equals zero.

Profit is revenue minus cost, so the difference rule gives the profit rate directly.

$$\frac{dP}{dq} = \frac{dR}{dq} - \frac{dC}{dq}$$

$$= \left(40 - 0.02q\right) - 12$$

$$= 28 - 0.02q$$

At $q = 1000$:

$$\frac{dP}{dq} = 28 - 20 = 8$$

At $q = 1500$:

$$\frac{dP}{dq} = 28 - 30 = -2$$

Setting $28 - 0.02q = 0$ gives $q = 1400$.

Final answer: profit rises by about $8$ dollars per extra item at $q = 1000$, falls by about $2$ dollars per item at $q = 1500$, and its rate is zero at $q = 1400$.

Why Does The Sum And Difference Rule Matter?

The rule matters because complicated functions are built from simple pieces, and it guarantees that their rates can be built the same way. Without it, every new polynomial or mixed function would need its own trip through the limit definition. Three places where it carries real weight:

• Economics. Profit is revenue minus cost, so marginal profit is marginal revenue minus marginal cost. Example 6 is the everyday version of this, and marginal cost and marginal revenue builds the full picture.

• Physics and flows. Net flow into a tank, net force on an object and net heat into a room are all sums and differences of separate contributions, so their rates of change are sums and differences of separate rates.

• Modelling. Scientists fit data with polynomials and sums of exponentials and sines precisely because linearity lets them differentiate and integrate those models one term at a time.

The history shows how special additivity is. In the same manuscript of 21 November 1675 where he first wrote the integral sign, Gottfried Leibniz also set down the product rule as a rule of its own; sums pass through a derivative untouched, but products never do. The statement of the sum rule and its limit proof appear in OpenStax Calculus Volume 1, Section 3.3 and, for general linear combinations, under linearity of differentiation.

Common Confusions With The Sum And Difference Rule

The rule itself is easy to state; most lost marks come from applying it where it does not belong or from losing a sign along the way.

Mistake 1: Extending The Rule To Products Or Quotients

Where it slips in:

Any expression with two factors, such as $(x^2 + 1)(x - 3)$ or $\frac{\sin x}{x}$, met right after a run of polynomial practice.

Don't do this:

Writing the derivative of $f \cdot g$ as $\frac{df}{dx} \cdot \frac{dg}{dx}$, or the derivative of $\frac{f}{g}$ as the quotient of the derivatives.

The correct way:

Use the product rule or the quotient rule, or rewrite the expression as a sum first when the algebra allows it, as in Example 3.

The rusher is the student most likely to do this, because after ten polynomials in a row "differentiate each piece" has become automatic and the multiplication sign goes unread.

Mistake 2: Dropping The Minus Sign On A Grouped Difference

Where it slips in:

Expressions such as $x^4 - (2x^2 + 5x)$, where a minus sign sits in front of a bracket.

Don't do this:

Writing $4x^3 - 4x + 5$, which applies the minus to the first term inside the bracket and forgets the second.

The correct way:

Distribute first: $x^4 - 2x^2 - 5x$, so the derivative is $4x^3 - 4x - 5$.

Students who are fluent with the derivative rules usually lose this mark to the algebra, not the calculus: the sign error happens one line before any differentiation starts. Writing the expanded expression out in full before differentiating is the habit that stops it.

Mistake 3: Adding Several Constants Of Integration

Where it slips in:

Integrating a sum term by term, as in $\int \left(3x^2 + 2\right) dx$.

Don't do this:

Writing $x^3 + C_1 + 2x + C_2$ and then treating $C_1$ and $C_2$ as two separate unknowns to solve for.

The correct way:

Combine them into one: $x^3 + 2x + C$. Two arbitrary constants added together are just one arbitrary constant.

The memorizer tends to write a $+,C$ after every single term, because "always add $C$" was learned as a ritual per integral sign rather than per answer.

Practice Problems On The Sum And Difference Rule

Work each problem before checking the answers below.

1. Differentiate $y = x^6 - 4x^3 + x - 9$.

2. Differentiate $y = 5\cos x + 2e^x - \sqrt{x}$ for $x > 0$.

3. Differentiate $y = \frac{x^3 - 2x + 1}{x^2}$ by splitting it into a sum first.

4. Evaluate $\int \left(3x^2 - 8x + 2\right) dx$.

5. Evaluate $\int_{0}^{\pi} \left(e^x + \cos x\right) dx$, exactly and to 4 decimal places.

6. Find the $x$-values where the curve $y = x^3 - 12x + 1$ has a horizontal tangent.

Answer to Problem 1: $6x^5 - 12x^2 + 1$.

Answer to Problem 2: $-5\sin x + 2e^x - \frac{1}{2\sqrt{x}}$.

Answer to Problem 3: after splitting, $y = x - 2x^{-1} + x^{-2}$, so the derivative is $1 + \frac{2}{x^2} - \frac{2}{x^3}$.

Answer to Problem 4: $x^3 - 4x^2 + 2x + C$.

Answer to Problem 5: $e^{\pi} - 1 \approx 22.1407$, since the cosine part contributes $\sin \pi - \sin 0 = 0$.

Answer to Problem 6: the derivative $3x^2 - 12$ equals $0$ at $x = -2$ and $x = 2$.

Wrapping Up

• The sum and difference rule lets you differentiate a sum or difference term by term, provided each term is differentiable at the point.

• Its proof is the limit definition plus the sum law for limits, so the rule is a consequence, not a convention.

• Geometrically, slopes add when graphs are added, and areas add when integrands are added.

• Combined with the constant multiple rule it gives linearity, which is why polynomials differentiate and integrate so easily.

• It does not extend to products or quotients, which need the product rule and quotient rule or a rewrite into a sum.

Your Next Move

Work through Practice Problems 1 to 6, then redo Example 3 without looking; if a product keeps getting differentiated as a product of derivatives, return to "Does The Sum And Difference Rule Work For Products And Quotients?", and if a sign goes missing in a bracket, return to Mistake 2.

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Read More

• Table of derivatives for the common functions

• Basic integration formulas for term-by-term integration

• The chain rule for composite functions

• The derivative of a function and what it measures

• How differentiation and integration undo each other