Power, Constant, Constant-Multiple, and Sum Rules

Four short rules — bring the exponent down, constants vanish, constant factors ride along, and sums split term by term — turn differentiating any polynomial into quick bookkeeping instead of a fresh limit calculation.

By the end you'll be able to differentiate any polynomial term by term with \(\frac{d}{dx}[x^n] = n x^{n-1}\) and its three companion rules — and explain where that shortcut comes from.

Predict: how does the slope at x = 1 change as you raise n from 1 to 5? Then drag the exponent slider and watch the tangent steepen.

This is \(f(x) = x^n\) with its dashed tangent line at the draggable point. The readout evaluates the power-rule formula \(n x^{n-1}\) at the same point — it always matches the tangent's measured slope, and the faint derivative curve \(f'(x) = n x^{n-1}\) passes through exactly that height. Hover the chart or the point for exact values.

x₀ = 1.00 · f(x₀) = 1.000 · tangent slope = 1·(1.00)0 = 1.000

f(x) = xⁿ — curve, tangent at x₀, and the derivative curve f′(x) = n·xⁿ⁻¹
f(x) = xⁿ tangent at x₀ f′(x) = n·xⁿ⁻¹

The power rule, constant rule, constant-multiple rule, and sum/difference rule combine to differentiate any polynomial — no limit computation required once you trust where they come from.

Formal

The four rules, where \(\frac{d}{dx}\) means "the derivative of": power rule \(\frac{d}{dx}[x^n] = n x^{n-1}\) — bring the exponent down as a factor and drop it by one. It holds for every real exponent \(n\), not just positive integers, so it also covers roots (\(x^{1/2}\)) and reciprocals (\(x^{-1}\)). Constant rule \(\frac{d}{dx}[c] = 0\) — a constant's graph is flat, so its slope is zero everywhere. Constant-multiple rule \(\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\) — a constant factor rides along untouched. And the sum/difference rule \(\frac{d}{dx}[f \pm g] = f' \pm g'\) — differentiate term by term.

Applied

This is why polynomials are the friendliest models in applied work: any quantity you model as a polynomial — a height, a cost, a population fit — hands you its rate of change by pure bookkeeping. A ball's height \(s(t) = -4.9t^2 + 20t + 2\) gives velocity \(s'(t) = -9.8t + 20\) in one line: power rule on \(t^2\), constant-multiple on the \(-4.9\) and the \(20\), constant rule on the \(+2\). No difference quotient, no limit — the rules did that work once, in general, so you never have to again.

Worked example

Differentiate \(f(x) = 4x^3 - 7x^2 + 5x - 9\). Go term by term (sum/difference rule), pulling each constant factor out front (constant-multiple rule) and applying the power rule: \(\frac{d}{dx}[4x^3] = 4 \cdot 3x^2 = 12x^2\); \(\frac{d}{dx}[-7x^2] = -7 \cdot 2x = -14x\); \(\frac{d}{dx}[5x] = 5 \cdot 1 = 5\); \(\frac{d}{dx}[-9] = 0\). So \(f'(x) = \) \(12x^2 - 14x + 5\).

Your turn

Differentiate \(g(x) = 2x^4 - 6x + 3\). Term by term: \(\frac{d}{dx}[2x^4] = 2 \cdot 4x^3 = 8x^3\); \(\frac{d}{dx}[-6x] = -6\); \(\frac{d}{dx}[3] = \) ____. Now put it together: \(g'(x) = \) ____

Reveal the answer

The lone constant differentiates to \(0\) (not 3, and not 1), so \(g'(x) = 8x^3 - 6 + 0 = \) \(8x^3 - 6\). Watch the two classic traps here: the \(-6x\) term is \(-6x^1\), whose derivative is \(-6 \cdot 1 \cdot x^0 = -6\) (not 0), while the bare \(+3\) vanishes entirely.

More info — why the power rule works (and what it replaces)

The power rule is a shortcut for the limit of the difference quotient — the limit you'd otherwise compute from scratch for every function. For a positive integer \(n\), expand \((x+h)^n\) with the binomial theorem: \((x+h)^n = x^n + n x^{n-1} h + (\text{terms with } h^2 \text{ and higher})\). Subtract \(x^n\), divide by \(h\), and let \(h \to 0\): every term except \(n x^{n-1}\) still carries a factor of \(h\) and vanishes, leaving exactly \(n x^{n-1}\). That's the same cancel-then-limit maneuver you used when computing derivatives from the definition — done once, in general, for all powers at the same time. The OpenStax chapter in Dive deeper below gives the full proofs, including the sum and constant-multiple rules.

Check your understanding

Question 1 of 4

Differentiate \(f(x) = 5x^3 + 2x^2 - 8x + 7\).

Question 2 of 4

What are \(\frac{d}{dx}[9]\) and \(\frac{d}{dx}[x]\), respectively?

Question 3 of 4

Differentiate \(y = \dfrac{1}{x^3}\). (Hint: rewrite it as a power of \(x\) first.)

Question 4 of 4

The power rule gives \(\frac{d}{dx}[x^2] = 2x\) instantly. What does the limit definition \(\displaystyle\lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}\) evaluate to, and why?

Recap

  • Power rule: \(\frac{d}{dx}[x^n] = n x^{n-1}\) for every real \(n\) — rewrite roots and reciprocals as powers (\(\sqrt{x} = x^{1/2}\), \(1/x^2 = x^{-2}\)) before applying it.
  • Constant rule: \(\frac{d}{dx}[c] = 0\); but \(x = x^1\) differentiates to \(1\), not 0.
  • Constant-multiple rule: \(\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\); sum/difference rule: \(\frac{d}{dx}[f \pm g] = f' \pm g'\).
  • Together they differentiate any polynomial term by term: \(4x^3 - 7x^2 + 5x - 9 \;\to\; 12x^2 - 14x + 5\).
  • The power rule is the once-and-for-all shortcut for the limit of the difference quotient — the binomial expansion shows every term but \(n x^{n-1}\) vanishes as \(h \to 0\).

Dive deeper

Sources

  • Power, Constant, Constant-Multiple, and Sum Rules