If f(x) = c (a constant), then f′(x) = 0 for every x. In notation: if f(x) = c, then df/dx = 0. Why: the slope of a horizontal line is 0. You can also see it from the limit definition: f′(x) = lim_{h→0} [f(x+h)-f(x)]/h = lim_{h→0} [c−c]/h = lim_{h→0} 0 = 0. This is one of the basic derivative rules in Topic 2.6 (constant rule) and is used with the sum, difference, constant multiple, and power rules to differentiate polynomials term-by-term (CED FUN-3.A.2–3.A.3). On the AP exam you’ll often use this to simplify expressions before applying other rules. For a quick review, see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm). For lots of practice problems, check https://library.fiveable.me/practice/ap-calculus.

Use linearity: derivative of a sum is the sum of derivatives, and a constant times x uses the constant multiple rule. For f(x) = 3x + 5: - d/dx[3x] = 3 · d/dx[x] = 3 · 1 = 3 (power rule / constant multiple) - d/dx[5] = 0 (constant function has zero derivative) So f′(x) = 3 + 0 = 3. Interpretation: the slope of the line 3x + 5 is 3 everywhere, so the instantaneous rate of change is constant 3. This aligns with CED FUN-3.A (sum and constant multiple rules and power rule). For a quick refresher, see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and try practice problems at (https://library.fiveable.me/practice/ap-calculus).

Use the sum rule when your function is written as a sum of two (or more) functions; use the difference rule when it’s written as a difference. Both come from the linearity of differentiation: d/dx [f(x) + g(x)] = f′(x) + g′(x) and d/dx [f(x) − g(x)] = f′(x) − g′(x). Practically that means you can differentiate term-by-term: for a polynomial like p(x) = 3x^4 − 5x^2 + 7, treat it as (3x^4) + (−5x^2) + 7 and apply constant-multiple and power rules to each term (the constant term has derivative 0). If you see a plus sign between parts, add the derivatives; if you see a minus sign, subtract the derivatives. This is exactly what the CED calls FUN-3.A.2/FUN-3.A.3 (term-by-term polynomial differentiation). For extra practice and AP-style problems, check the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and the Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2).

Short answer: they’re different rules for different situations. - Constant rule: if f(x) = C (a constant number), then f′(x) = 0 for all x. Example: if f(x) = 7, f′(x) = 0. This follows from the limit definition and is tested on the AP as "zero derivative" / constant function. - Constant multiple rule: if g(x) = k·f(x) where k is a constant and f is differentiable, then g′(x) = k·f′(x). Example: if g(x) = 5x^3, then g′(x) = 5·3x^2 = 15x^2. This is part of the linearity of differentiation (sum/difference + constant multiple) in the CED (FUN-3.A.2, FUN-3.A.3). Use them together when needed: derivative of a polynomial term a·x^n uses the constant multiple rule with the power rule; the derivative of a standalone number uses the constant rule. For more practice and AP-aligned notes see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2).

Start with the rules you already know from Topic 2.6: differentiation is linear (sum/difference and constant multiple rules) and use the power rule: d/dx[x^n] = n x^(n-1). Now do it term-by-term. f(x) = 2x^3 − 4x + 7 - For 2x^3: constant multiple rule → 2 · d/dx[x^3]. Power rule gives d/dx[x^3] = 3x^2, so this term’s derivative is 2 · 3x^2 = 6x^2. - For −4x: treat as (−4)·x. d/dx[x] = 1, so derivative is −4 · 1 = −4. - For +7: constant function → derivative is 0. Combine: f′(x) = 6x^2 − 4. That’s it—polynomial differentiation is just term-by-term using linearity and the power rule. If you want more worked examples and AP-aligned practice, check the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and thousands of practice problems (https://library.fiveable.me/practice/ap-calculus).

Use the constant multiple rule together with the power rule any time a term is a constant times a power of x. Differentiation is linear: d/dx[c·f(x)] = c·d/dx[f(x)], and the power rule says d/dx[x^n] = n x^(n−1). So for a term like 5x^3, do both rules: d/dx[5x^3] = 5·d/dx[x^3] = 5·3x^2 = 15x^2. For (1/2)x^(−1): d/dx[(1/2)x^(−1)] = (1/2)(−1)x^(−2) = −(1/2)x^(−2). For polynomials, differentiate term-by-term: d/dx[3x^4 − 7x + 2] = 12x^3 − 7 + 0 (constant term → 0), which is exactly the FUN-3.A idea in the CED. Quick cautions: if the constant multiplies a non-power expression (like (x+2)^3 or sin x), you still use the constant multiple rule but may also need chain/product rules (those are other topics). For AP review and more examples, see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2). Practice lots of term-by-term problems at (https://library.fiveable.me/practice/ap-calculus).

You always start by using the linearity of differentiation: the derivative of a sum/difference is the sum/difference of the derivatives, and constants pull out. So treat the function term-by-term. Quick checklist: - If you have a sum or difference, split it and differentiate each term separately (sum/difference rule). - If a term has a constant factor, pull it out and apply the constant multiple rule. - For power-like terms (x^n), apply the power rule next. - Constants differentiate to 0. - If a term is more complicated (product, quotient, composite), you’ll need product/quotient/chain rules—handle those inside the term after using linearity to separate the expression. Example: f(x)=3x^4 − 5x + 7 Step 1 (sum/diff): differentiate each term. Step 2 (const multiple & power): 3·d/dx(x^4) − 5·d/dx(x) + d/dx(7) Step 3 (power/constant): 3·4x^3 − 5·1 + 0 = 12x^3 − 5 On the AP exam you’re expected to apply these term-by-term rules for polynomials and familiar functions (FUN-3.A.2/3.A.3). For more practice see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

Use the power rule term-by-term (and the sum/difference & constant multiple rules). For f(x) = 5x^4 − 3x^2 + 8x − 1, f′(x) = 5·4x^3 − 3·2x + 8·1 − 0 = 20x^3 − 6x + 8. So the derivative is 20x^3 − 6x + 8. This is exactly an application of the power rule and linearity of differentiation (FUN-3.A.2–3 in the CED). If you want a quick review of these rules, see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm). For extra practice problems, check the unit practice page (https://library.fiveable.me/practice/ap-calculus).

Think of the derivative as the slope of the graph (instantaneous rate of change). A constant function f(x) = c is a horizontal line at height c, so its slope is 0 everywhere—nothing is changing. Using the limit definition: f′(x) = lim(h→0) [f(x+h) − f(x)]/h = lim(h→0) [c − c]/h = lim(h→0) 0/h = 0. That calculation matches the rules in Topic 2.6: constants differentiate to zero, and differentiation is linear (sum, difference, constant multiple). So when you differentiate a polynomial term-by-term (power rule + sum rule), any constant term just drops out because its derivative is 0—that’s why derivatives give you the rate-of-change parts only. This is exactly the kind of basic fact the AP exam expects you to use (CED FUN-3.A rules). For a quick refresher, see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and more practice at (https://library.fiveable.me/practice/ap-calculus).

You use the linearity of differentiation: derivative of a constant is 0, and d/dx of a sum/difference or constant multiple is done term-by-term. In symbols: - d/dx[c] = 0 - d/dx[cf(x)] = c·f′(x) - d/dx[f(x) ± g(x)] = f′(x) ± g′(x) So first rewrite the function as a sum of terms (monomials + constants), then apply the power rule to each term and keep signs and coefficients. Example: f(x) = 3x^4 − 5x^2 + 7x − 4 f′(x) = 3·4x^3 − 5·2x + 7·1 − 0 = 12x^3 − 10x + 7 For polynomials this is exactly what the CED expects (FUN-3.A.2/FUN-3.A.3): differentiate term-by-term using constant, sum/difference, and constant multiple rules combined with the power rule. Want more worked examples and practice problems? See the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm), the Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2), and lots of practice questions (https://library.fiveable.me/practice/ap-calculus).

Think of differentiation as linear—you can take derivatives term-by-term. The power rule says d/dx[x^n] = n x^(n-1). The sum/difference and constant-multiple rules let you apply that to each term and pull constants out. Quick steps: - If f(x) = a·g(x), then f'(x) = a·g'(x). (constant multiple) - If F(x) = g(x) ± h(x), then F'(x) = g'(x) ± h'(x). (sum/difference) Example: f(x) = 3x^4 − 5x^2 + 7x − 9 - Differentiate each term with power rule: d/dx[3x^4] = 3·4x^3 = 12x^3 d/dx[−5x^2] = −5·2x = −10x d/dx[7x] = 7 d/dx[−9] = 0 So f'(x) = 12x^3 − 10x + 7. This term-by-term process is exactly what's tested in Topic 2.6 (FUN-3.A.2/3.A.3). For more examples and practice, see the topic study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and try problems on Fiveable Practice (https://library.fiveable.me/practice/ap-calculus).

Use the constant-multiple rule + power rule. For f(x) = −6x^2 you can pull out the constant and differentiate x^2: f'(x) = −6 · d/dx(x^2) = −6 · (2x) = −12x. So the derivative is −12x (or dy/dx = −12x). This uses linearity of differentiation (constant multiple) and the power rule (d/dx x^n = n x^{n−1}), exactly the FUN-3.A ideas in the CED for Topic 2.6. On the AP exam you’ll often apply these rules term-by-term for polynomials, e.g., d/dx(3x^3 − 5x + 7) = 9x^2 − 5 + 0. If you want a short review of these rules and practice, see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and try some practice problems (https://library.fiveable.me/practice/ap-calculus).

Start by using linearity of the derivative (sum/difference and constant multiple rules) and the power rule. Let f(x) = 4x^3 + 2x − 9. 1) Differentiate term-by-term: f′(x) = d/dx[4x^3] + d/dx[2x] + d/dx[−9]. 2) Apply constant multiple + power rule to 4x^3: d/dx[4x^3] = 4·d/dx[x^3] = 4·(3x^2) = 12x^2. 3) For 2x (power rule with n = 1): d/dx[2x] = 2·d/dx[x] = 2·1 = 2. 4) Constant rule for −9: d/dx[−9] = 0. Combine: f′(x) = 12x^2 + 2. So dy/dx = 12x^2 + 2 (this uses the power rule plus sum/difference and constant-multiple properties from the CED). For more practice and review on these rules, see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and try extra problems at (https://library.fiveable.me/practice/ap-calculus).

Yes—you take the derivative term-by-term. Differentiation is linear: the derivative of a sum/difference is the sum/difference of the derivatives, and constants pull out. Use the power rule for each term. For f(x) = 3x^2 + 5x − 2: - d/dx[3x^2] = 3 · d/dx[x^2] = 3 · (2x) = 6x (constant multiple + power rule) - d/dx[5x] = 5 · d/dx[x] = 5 · 1 = 5 - d/dx[−2] = 0 (constant function) So f′(x) = 6x + 5. This directly matches CED FUN-3.A (power rule + sum/difference/constant-multiple rules) and is exactly the kind of polynomial work seen on the AP exam. For more examples and practice, see the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and the unit page (https://library.fiveable.me/ap-calculus/unit-2).

Short answer: yes—the constant multiple rule says d/dx[c·f(x)] = c·f '(x). Differentiation is linear, so constants factor out. That’s why you can do term-by-term derivatives of polynomials (power rule + constant multiple). Quick examples: - d/dx[5x^3] = 5·d/dx[x^3] = 5·3x^2 = 15x^2. - d/dx[7·x] = 7·1 = 7. - d/dx[7] = 0 (constant function → zero derivative). Common slip: don’t confuse a constant factor outside the function with a constant added inside. For c·f(x) you just pull c out; for f(cx) or f(x)+c you may need the chain rule or nothing to do with constant multiple. On the AP, use this with the power rule and sum/difference rules to handle polynomials and coefficients term-by-term (CED FUN-3.A.2/3.A.3). For a quick refresher and practice, check the Topic 2.6 study guide (https://library.fiveable.me/ap-calculus/unit-2/derivative-rules-constant-sum-difference-constant-multiple/study-guide/0DS7QaXV5BZFYbNZySlm) and try problems from the Unit 2 page (https://library.fiveable.me/ap-calculus/unit-2) or the practice bank (https://library.fiveable.me/practice/ap-calculus).