The power rule and basic differentiation rules

When you reach for it

Finding a derivative from the definition works for any function, but each one costs a page of algebra: expand, cancel, let h \to 0. The answers it gave already rhyme. The derivative of x^2 is 2x, of x^3 is 3x^2, of x^4 is 4x^3. The exponent comes down in front and drops by one.

This lesson proves that pattern, the power rule, together with three rules for building functions out of pieces: the constant rule, the constant multiple rule and the sum and difference rule. Each is proved once from the limit definition

and after that the limit never has to be taken again for these functions. Reach for these rules whenever a function is a sum of constant multiples of powers of x: every polynomial, and, after a rewrite, many expressions with roots and fractions.

The lesson runs over three days. Part 1 proves the four rules. Part 2 differentiates polynomials and rewrites expressions so the rules apply. Part 3 extends the power rule to negative and fractional exponents and finds where it stops.

The procedure

Part 1: the rules, each proved from the definition

The constant rule. If f(x) = c for a constant c, then f(x + h) - f(x) = c - c = 0, so the difference quotient is 0 for every h \ne 0 and its limit is 0:

The graph of a constant is a horizontal line, and a horizontal line has slope 0 everywhere.

The constant multiple rule. Let k be a constant and let f be differentiable at x. The difference quotient of kf(x) is

The quotient on the right has limit f'(x), and by the constant multiple law of the limit laws the limit of k times it is k times that limit:

A vertical stretch by k multiplies every rise by k and leaves every run alone, so every slope is multiplied by k.

The sum and difference rule. Let f and g both be differentiable at x. Regroup the quotient of f + g, using the associative and commutative properties of addition:

Both quotients on the right have limits, f'(x) and g'(x), so the sum law says the limit of their sum is the sum of the limits. A difference f - g is f + (-1)g, so the constant multiple rule with k = -1 finishes it:

Applied again and again, the rule covers a sum of any number of terms.

The power rule for positive integers. Let n be a positive integer. By the binomial theorem,

The second coefficient is \binom{n}{1} = n, the number of ways to pick the single factor (x + h) that contributes h while the other n - 1 factors contribute x (a count from Counting). Subtract x^n: the first term cancels and every term left carries at least one h. Divide by h, allowed because h \ne 0:

Every term after the first still carries a factor of h, so the right side is a polynomial in h whose value at h = 0 is n x^{n-1}. Its limit is that value:

For n = 1 this says \displaystyle \frac{d}{dx}[x] = 1: the line y = x has slope 1. Together, the four rules confirm a result found the long way: the definition gave \displaystyle \frac{d}{dx}\big[2x^2 + 5x\big] = 4x + 5, and the rules give 2(2x) + 5(1) = 4x + 5 in one line.

Part 2: polynomials, rewriting, and notation

A polynomial is a sum of constant multiples of positive integer powers of x, plus a constant. The sum rule splits it into terms, the constant multiple rule pulls out each coefficient, the power rule handles each power, and the constant rule sends the constant term to 0. So the derivative of a polynomial of degree n \ge 1 is a polynomial of degree n - 1.

The rules say nothing yet about a product or a quotient of two functions. In particular the derivative of a product is not the product of the derivatives (Example 4 shows a case). Until the product rule arrives, a product of polynomials is expanded first, and a quotient whose bottom is a single power of x is split term by term. A rewrite changes the form of the function and not its values, so the derivative is unchanged, except that a rewrite which divides by x is valid only for x \ne 0.

Two notations run side by side. For y = f(x), prime notation writes the derivative f'(x) and its value at a as f'(a). Leibniz notation writes the same function \dfrac{dy}{dx} or \dfrac{d}{dx}\big[f(x)\big], and its value at a as \left.\dfrac{dy}{dx}\right|_{x = a}. Leibniz notation names the input variable, which matters when the variable is t or u: \dfrac{d}{dt}\big[t^3\big] = 3t^2. The symbol \dfrac{dy}{dx} is read as one symbol, the limit of \dfrac{\Delta y}{\Delta x}, not as a fraction of two numbers.

Part 3: negative and fractional exponents

The power rule keeps its form for other exponents:

at every x where both x^r and x^{r-1} are defined, as long as x^r is defined on both sides of that x. The cases are not all proved by the same tool, and it is worth saying exactly which are proved here.

Negative integers, proved here. Let n be a positive integer and f(x) = x^{-n} = \dfrac{1}{x^n} with x \ne 0. Over the common denominator x^n(x + h)^n,

From Part 1, \dfrac{(x + h)^n - x^n}{h} tends to n x^{n-1}, so the top over h tends to -n x^{n-1}. The bottom x^n(x + h)^n tends to x^{2n}, which is not 0, so the quotient law applies:

That is the power rule with r = -n. The quotient rule will give it again in one line.

Unit fractions, proved here. Let q \ge 2 be an integer and f(x) = x^{1/q}, the q-th root from rational exponents, with x > 0. Write a = (x + h)^{1/q} and b = x^{1/q}, so a^q - b^q = h. The factoring a^q - b^q = (a - b)(a^{q-1} + a^{q-2}b + \cdots + b^{q-1}) gives

As h \to 0, a \to b because a root is continuous, so each of the q terms in the bottom tends to b^{q-1}:

For odd q the same argument works for x < 0 too.

Other rational exponents, stated here and proved later. For \displaystyle r = \frac{p}{q} in general, x^{p/q} = \left(x^{1/q}\right)^p is a power of a root, and differentiating a function inside another function is the job of the chain rule; implicit differentiation gives a second proof. This lesson uses the rule for every rational exponent and marks that it rests on those later proofs. Irrational exponents such as x^{\pi} need logarithms and wait for the derivatives of exponential and log functions.

To use the rule on a root or a fraction, first rewrite it as a power of x: \displaystyle \sqrt[q]{x^p} = x^{p/q} and \dfrac{1}{x^r} = x^{-r}. A coefficient stays outside as a constant multiple: \dfrac{5}{x^2} = 5x^{-2}.

Worked examples

Boundaries

• The sum and constant multiple rules need the pieces to be differentiable. \displaystyle \frac{d}{dx}\big[f + g\big] = f' + g' is proved only where f'(x) and g'(x) both exist. The limit laws behind it need both limits to exist.
• The power rule is about a power of the variable. In x^r the base varies and the exponent r is a constant. It says nothing about a constant base raised to a variable, such as 2^x, and a constant such as \pi^2 has derivative 0.
• The rules cover sums and constant multiples only. A product or quotient of two non-constant functions is not differentiated piece by piece: rewrite it first, or wait for the product and quotient rules.
• Fractional and negative powers have a restricted domain. x^{-n} and its derivative need x \ne 0. An even root such as x^{1/2} needs x \ge 0, and at x = 0 it has no derivative, since there is no room to its left and the one-sided quotient \displaystyle \frac{\sqrt{h}}{h} grows without bound. At x = 0 a power x^{p/q} with \displaystyle 0 < \frac{p}{q} < 1 has no derivative: the graph has a cusp (x^{2/3}) or a vertical tangent (x^{1/3}). When \displaystyle \frac{p}{q} > 1 and x^{p/q} is defined on both sides of 0, as for x^{4/3}, the derivative at 0 exists and is 0.
• The rewrite must be valid. Dividing through by x holds only for x \ne 0, and the derivative inherits that restriction.

Where this goes next

With these rules, every polynomial is differentiated by inspection, and roots and reciprocals follow after a rewrite. The product rule and the quotient rule remove the need to expand or split, and the chain rule finishes the proof of the power rule for every rational exponent. Second derivatives apply the same rules twice. Running the power rule backward, raising the exponent and dividing, is finding an antiderivative. Setting a polynomial’s derivative equal to 0 locates the flat points of its graph, where maxima and minima are found.

Exercises

Part 1: the rules

Find each derivative.

1. f(x) = x^{12}
2. \displaystyle g(x) = -\frac{2}{5}x^{10}
3. \displaystyle k(x) = 8\sqrt{3}
4. \dfrac{d}{du}\left[\tfrac{1}{4}u^8\right]
5. The functions f and g are differentiable with f'(3) = -2 and g'(3) = 5. Find h'(3) for h(x) = 2f(x) - 3g(x) + x^4.

Part 2: polynomials and rewriting

Find the derivative. Where a rewrite divides by x, say for which x the answer holds.

6. p(x) = 2x^6 - 5x^3 + x - 11
7. y = \dfrac{x^4}{4} - \dfrac{x^3}{3} + \dfrac{x^2}{2} - x (give \dfrac{dy}{dx})
8. s(t) = -4.9t^2 + 14t + 2 (give \dfrac{ds}{dt})
9. q(x) = (3x - 2)^2
10. r(x) = x^2(x - 5)(x + 1)
11. w(x) = \dfrac{8x^5 - 6x^3 + x^2}{2x^2}
12. For f(x) = 2x^3 + 3x^2 - 36x + 5, find f'(0) and every x at which f'(x) = 0.
13. Nurgul wants the curve y = ax^3 - 2x to have slope 10 at x = 2. Find the constant a.

Cholpon sews felt rugs in her workshop. Making x rugs in a month costs C(x) = 0.03x^3 - 0.9x^2 + 20x + 150 dollars.

14. Find C'(x) and C'(10), and give the units of C'(10).
15. For which positive number of rugs is C'(x) = 20?

Part 3: negative and fractional exponents

Find the derivative, and say for which x it holds.

16. f(x) = x^{-5}
17. g(x) = \dfrac{3}{x^4} - \dfrac{2}{x}
18. \displaystyle y = \sqrt[4]{x^3}
19. \displaystyle h(x) = 10\sqrt{x} - \dfrac{6}{\sqrt[3]{x}}
20. \displaystyle k(x) = \dfrac{x^2 + 4}{\sqrt{x}}; also find k'(4).
21. Use the difference quotient at 0 to decide whether f(x) = x^{4/3} and g(x) = x^{1/3} have a derivative at x = 0.

Reasoning

22. Show that \dfrac{d}{dx}\big[f(x) - g(x)\big] = f'(x) - g'(x) using only the sum rule and the constant multiple rule.
23. Use the binomial theorem to write \dfrac{(x + h)^6 - x^6}{h} as a polynomial in h, and explain why its limit as h \to 0 is 6x^5.
24. Aibek says the derivative of x^2 \cdot x^3 is 2x \cdot 3x^2 = 6x^3. Find the correct derivative by rewriting, and show his answer is wrong at x = 1.
25. The proof in Part 1 covers only positive integer exponents. Explain where it uses that the exponent is a positive integer, and say which tools give the power rule for negative integers, for \displaystyle \frac{1}{q}, and for a general rational exponent.