Derivative sum rule: Distribute differentiation over addition

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | FAQ

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The Principle

The move: Distribute $\frac{d}{dx}$ over a sum — replace the derivative of a sum with the sum of the separate derivatives.

The invariant: This produces an equivalent derivative expression at any point where both $f$ and $g$ are differentiable.

Pattern: $\frac{d}{dx}(f(x)+g(x)) \quad\longrightarrow\quad f'(x)+g'(x)$

Legal ✓	Illegal ✗
$\frac{d}{dx}(x^2 + e^x) \to 2x + e^x$; both differentiable everywhere	$\frac{d}{dx}(\lvert x\rvert + (-\lvert x\rvert))\big\rvert_{x=0} \not\to \frac{d}{dx}\lvert x\rvert\big\rvert_{x=0} + \frac{d}{dx}(-\lvert x\rvert)\big\rvert_{x=0}$; neither component differentiable at $x = 0$

In the Illegal column: the sum $|x| + (-|x|) = 0$ is the constant function zero — differentiable everywhere, including at $x = 0$ with derivative $0$. But $\frac{d}{dx}|x|$ does not exist at $x = 0$ (left-hand derivative $-1$, right-hand derivative $+1$), so the condition “f and g differentiable” fails for both components. A differentiable sum does not authorize the split — only individually differentiable components do.

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Conditions of Applicability

Condition: f and g differentiable

Both $f$ and $g$ must be differentiable at the point (or on the interval) where you apply the rule. A function can be continuous everywhere but fail to be differentiable at an isolated point — $|x|$ at $x = 0$ is the standard example — so continuity alone does not authorize the split.

Before applying, check: confirm that both $f(x)$ and $g(x)$ are differentiable at the evaluation point.

• When either term in the sum lacks a derivative at the evaluation point, splitting $\frac{d}{dx}$ references a value that does not exist — the move is not valid there.
• Polynomials, $e^x$, $\sin x$, $\cos x$, and power functions $x^n$ with integer $n$ (away from $x = 0$ for $n < 0$) are differentiable on their standard domains; the condition holds automatically for sums of these functions.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: apply the derivative sum rule to $\frac{d}{dx}(f(x) \cdot g(x))$ — treating multiplication like addition — and write $f'(x) + g'(x)$ instead of the correct product-rule result $f'(x)g(x) + f(x)g'(x)$.

Debug: check whether the two expressions are joined by $+$ (sum rule: split cleanly) or $\times$ (product rule: cross-multiply). Differentiation distributes over addition, not multiplication.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• The condition says “f and g differentiable” — does it require differentiability everywhere on $\mathbb{R}$, or just at the point of evaluation? What if one term is differentiable on $(0, \infty)$ but not at $x = 0$?
• Differentiation distributes over addition. Does it also distribute over multiplication? Why or why not — and what rule handles the multiplication case?

For the Principle

• When differentiating a polynomial $a_n x^n + \cdots + a_1 x + a_0$, how many times would you apply the sum rule, and why does the condition always hold for polynomial terms?
• If $f$ is differentiable and $g$ is continuous but not differentiable at a point, can you apply the sum rule there? What does your answer reveal about the importance of the individual-term check?

Between Principles

• The derivative sum rule and the power rule are applied together in almost every polynomial derivative. In what order do the two rules appear in the chain, and why does the sum rule come first?

Generate an Example

• Construct a sum $f(x) + g(x)$ where the combined sum is differentiable at every real number, but at least one of $f$ and $g$ is not differentiable at some point. What does this tell you about the direction in which the sum rule can operate?

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Distribute d/dx over a sum: the derivative of f(x)+g(x) is f'(x)+g'(x).

Write the canonical equation: _____$\frac{d}{dx}(f(x)+g(x)) = f'(x)+g'(x)$

State the canonical condition: _____f and g differentiable

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to use worked examples effectively.)

Procedure Walkthrough

Starting from $\frac{d}{dx}(x^4 + \sin x)$, reach the evaluated derivative.

Step	Expression	Operation
0	$\frac{d}{dx}(x^4 + \sin x)$	—
1	$\frac{d}{dx}(x^4) + \frac{d}{dx}(\sin x)$	Derivative sum rule — both $x^4$ and $\sin x$ are differentiable everywhere; condition satisfied
2	$4x^{4-1} + \frac{d}{dx}(\sin x)$	Power rule on $x^4$: $n = 4 \in \mathbb{Z}$ ✓, bring exponent down as coefficient
3	$4x^3 + \cos x$	Arithmetic: $4 - 1 = 3$; standard result $\frac{d}{dx}\sin x = \cos x$

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Drills

Action label: Identify the rule applied

What rule was used between these two states?

$\frac{d}{dx}(x^3 + e^x) \quad\longrightarrow\quad \frac{d}{dx}(x^3) + \frac{d}{dx}(e^x)$

Reveal

Derivative sum rule — a single derivative of a sum was split into the sum of two separate derivatives. Both $x^3$ and $e^x$ are differentiable everywhere, so the condition holds.

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What rule was used between these two states?

$\frac{d}{dx}(5x^2 + \ln x) \quad\longrightarrow\quad \frac{d}{dx}(5x^2) + \frac{d}{dx}(\ln x) \quad (x > 0)$

Reveal

Derivative sum rule. Both $5x^2$ and $\ln x$ are differentiable on $(0, \infty)$, so the split is valid for $x > 0$.

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[Near-miss — negative] Is this step valid? Explain.

$\frac{d}{dx}\!\left(\lvert x\rvert + (-\lvert x\rvert)\right)\!\bigg|_{x=0} \quad\longrightarrow\quad \left.\frac{d}{dx}\lvert x\rvert\right|_{x=0} + \left.\frac{d}{dx}(-\lvert x\rvert)\right|_{x=0}$

Reveal

Invalid. The sum $|x| + (-|x|) = 0$ is the constant function zero — differentiable everywhere, with derivative $0$ at $x = 0$. But $\frac{d}{dx}|x|$ does not exist at $x = 0$: the left-hand derivative is $-1$ and the right-hand derivative is $+1$, so $|x|$ has a corner there. The condition “f and g differentiable” fails for both components, so the split cannot be applied at that point. This is the canonical near-miss: a differentiable sum does not authorize the split — only individually differentiable components do.

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[Negative] Which of these derivative sum rule applications is valid at $x = 0$? Give a one-line reason for each.

(a) $\frac{d}{dx}(x^2 + \cos x)\big|_{x=0}$

(b) $\frac{d}{dx}(\lvert x\rvert + x^3)\big|_{x=0}$

(c) $\frac{d}{dx}(e^x + \sqrt{x})\big|_{x=0}$

Reveal

(a) Valid — $x^2$ and $\cos x$ are both differentiable at $x = 0$.

(b) Invalid — $\lvert x\rvert$ is not differentiable at $x = 0$; the condition fails for the first term.

(c) Invalid — $\sqrt{x} = x^{1/2}$ is not differentiable at $x = 0$ (the derivative $\frac{1}{2\sqrt{x}}$ diverges as $x \to 0^+$); the condition fails for the second term.

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Forward step: Apply the rule

Apply the derivative sum rule, then differentiate each term.

$\frac{d}{dx}(x^3 + 4x)$

Reveal

$\frac{d}{dx}(x^3) + \frac{d}{dx}(4x) = 3x^2 + 4$

Both terms are differentiable everywhere; condition holds.

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Apply the derivative sum rule, then differentiate each term.

$\frac{d}{dx}(e^x + \cos x)$

Reveal

$\frac{d}{dx}(e^x) + \frac{d}{dx}(\cos x) = e^x - \sin x$

Both $e^x$ and $\cos x$ are differentiable everywhere.

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Apply the derivative sum rule to a three-term sum (apply twice).

$\frac{d}{dx}(x^2 + \sin x + e^x)$

Reveal

First application: $\frac{d}{dx}(x^2 + \sin x) + \frac{d}{dx}(e^x)$

Second application: $\frac{d}{dx}(x^2) + \frac{d}{dx}(\sin x) + \frac{d}{dx}(e^x)$

Evaluate: $2x + \cos x + e^x$

Each application is valid — all three functions are differentiable everywhere.

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Apply the derivative sum rule, then differentiate each term (restrict to $x > 0$).

$\frac{d}{dx}(\ln x + x^4)$

Reveal

$\frac{d}{dx}(\ln x) + \frac{d}{dx}(x^4) = \frac{1}{x} + 4x^3 \quad (x > 0)$

Both $\ln x$ and $x^4$ are differentiable on $(0, \infty)$; condition holds for all $x > 0$.

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Apply the derivative sum rule, then differentiate each term ($x \neq 0$).

$\frac{d}{dx}(x^5 + x^{-2})$

Reveal

$\frac{d}{dx}(x^5) + \frac{d}{dx}(x^{-2}) = 5x^4 + (-2)x^{-3} = 5x^4 - \frac{2}{x^3} \quad (x \neq 0)$

Both $x^5$ and $x^{-2}$ are differentiable on $x \neq 0$.

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Transition identification: Locate the rule in a chain

The following evaluation has three transitions. Which transition uses the derivative sum rule?

Transition	From	To
0→1	$\frac{d}{dx}(3x^2 + \sin x)$	$\frac{d}{dx}(3x^2) + \frac{d}{dx}(\sin x)$
1→2	$\frac{d}{dx}(3x^2) + \frac{d}{dx}(\sin x)$	$3\frac{d}{dx}(x^2) + \cos x$
2→3	$3\frac{d}{dx}(x^2) + \cos x$	$6x + \cos x$

Reveal

Transition 0→1 uses the derivative sum rule — a single derivative of a sum is split into two separate derivatives.

Transition 1→2 applies the derivative constant multiple rule to $3x^2$ (pulling the factor $3$ outside) and the standard derivative $\frac{d}{dx}\sin x = \cos x$.

Transition 2→3 applies the power rule: $\frac{d}{dx}(x^2) = 2x$, so $3 \cdot 2x = 6x$.

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In this chain, identify which step uses the derivative sum rule and which uses the power rule.

$\frac{d}{dx}(x^3 + x) \;\longrightarrow\; \frac{d}{dx}(x^3) + \frac{d}{dx}(x) \;\longrightarrow\; 3x^2 + 1$

Reveal

• Step 0→1: derivative sum rule — splits the derivative of a sum into two separate derivatives.
• Step 1→2: power rule — $\frac{d}{dx}(x^3) = 3x^2$ and $\frac{d}{dx}(x) = x^1 \to 1 \cdot x^0 = 1$.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: Starting from $\frac{d}{dx}(2x^3 + \cos x + e^x)$, find the derivative using the derivative sum rule.

Full solution

Step	Expression	Move
0	$\frac{d}{dx}(2x^3 + \cos x + e^x)$	—
1	$\frac{d}{dx}(2x^3 + \cos x) + \frac{d}{dx}(e^x)$	Derivative sum rule — all three terms differentiable; condition satisfied
2	$\frac{d}{dx}(2x^3) + \frac{d}{dx}(\cos x) + \frac{d}{dx}(e^x)$	Derivative sum rule again
3	$6x^2 + (-\sin x) + e^x$	Constant multiple + power rule; $\frac{d}{dx}\cos x = -\sin x$; $\frac{d}{dx}e^x = e^x$
4	$6x^2 - \sin x + e^x$	Collect signs

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Related Principles

Principle	Relationship
Derivative at a point	Foundation — the sum rule is a reusable consequence of the derivative definition, and it only makes sense after the derivative object is established
Power rule	Applied immediately after the sum rule in polynomial and mixed-term derivatives; the sum rule splits, then the power rule differentiates each term
Derivative constant multiple rule	Companion to the sum rule — together they handle any linear combination $c_1 f + c_2 g$ without needing the product rule
Product rule	Handles $\frac{d}{dx}(fg)$; the sum rule does not distribute over multiplication, so these two rules are complementary entry points depending on whether terms are added or multiplied

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FAQ

What is the derivative sum rule?

The derivative sum rule states that differentiation distributes over addition: if $f$ and $g$ are both differentiable at the evaluation point, then $\frac{d}{dx}(f(x)+g(x)) = f'(x)+g'(x)$. The rule authorizes replacing the derivative of a sum with the sum of the separate derivatives whenever both component functions are differentiable there.

When does the derivative sum rule apply?

It applies when both $f$ and $g$ are differentiable at the evaluation point. Polynomials, $e^x$, $\sin x$, $\cos x$, and $\ln x$ (on $x > 0$) are differentiable on their standard domains. Functions with cusps, corners, or vertical tangents — such as $|x|$ at $x = 0$ or $x^{1/2}$ at $x = 0$ — are not differentiable at those specific points, so the rule cannot be applied there even though the sum may be differentiable.

Does the derivative sum rule work for three or more terms?

Yes. Apply it repeatedly: split a three-term sum by using the rule twice, a four-term sum three times, and so on. For any polynomial $a_n x^n + \cdots + a_0$, the condition holds everywhere, so you can apply the rule repeatedly across all terms. Each individual application of the rule is independently valid.

How is the derivative sum rule different from the product rule?

The sum rule distributes cleanly: $\frac{d}{dx}(f+g) = f' + g'$. The product rule does not: $\frac{d}{dx}(fg) = f'g + fg'$ requires a cross-multiply form. Writing $\frac{d}{dx}(fg) = f' + g'$ (applying the sum rule to a product) is a common misapplication that produces a wrong result in almost every case. Check whether the operation between the two expressions is $+$ (sum rule) or $\times$ (product rule) before choosing.

What happens when one term in the sum is not differentiable at the evaluation point?

The split cannot be applied at that point — distributing $\frac{d}{dx}$ would reference a component derivative that does not exist. Note that the sum $f + g$ might still be differentiable at that point (as with $|x| + (-|x|) = 0$), but the rule cannot be used to find that derivative by splitting. You would need a different approach, such as the limit definition.

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How This Fits in Unisium

The derivative sum rule is a move-selection exercise as much as a mechanical step. The fluency built here — recognizing that addition allows splitting, confirming differentiability before applying the rule, and distinguishing sum-rule transitions from product-rule transitions in a worked chain — transfers to every multi-term derivative in calculus. Unisium trains this pattern with state-transition drills so that condition checking becomes automatic rather than something you have to consciously remember under exam pressure.

Explore further:

• Calculus Subdomain Map — Return to the calculus hub to see how the linearity rules and structural rules fit into the same derivative cluster
• Power rule — The rule most commonly paired with the sum rule when differentiating polynomial and mixed expressions
• Elaborative Encoding — Build deep understanding of why “f and g differentiable” is a real constraint, not a formality
• Retrieval Practice — Make the equation and condition retrievable without effort

Ready to master the derivative sum rule? Start practicing with Unisium or explore the full learning framework in Masterful Learning.

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