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Unit 2: Differentiation: Definition and Fundamental Properties

Derivative Rules: Constant, Sum, Difference, and Constant Multiple

Differentiate any polynomial, and any sum of powers, term by term, and see why constants vanish while constant multipliers stay.

Topic 2.6 · 7 min read

Finding the derivative of p(x)=4x3−7x+2p(x) = 4x^3 - 7x + 2 from the limit definition means expanding (x+h)3(x+h)^3, cancelling, and dividing by hh, and the next polynomial would need the same work again. Instead, you can differentiate 4x34x^3, −7x-7x and 22 separately and combine the results. Three rules make that legal: a constant has derivative 0, a constant multiplier passes through the derivative, and the derivative of a sum or difference is the sum or difference of the derivatives. Each is a short consequence of the limit definition, and together with the power rule (topic 2.5) they differentiate every polynomial.

A constant has derivative zero

The graph of a constant function f(x)=cf(x) = c is a horizontal line, and a horizontal line has slope 0 everywhere. The limit definition says the same thing: f(x+h)f(x+h) and f(x)f(x) are both cc, so every difference quotient is

f(x+h)−f(x)h=c−ch=0,\frac{f(x+h) - f(x)}{h} = \frac{c - c}{h} = 0,

and the limit as h→0h \to 0 is 0.

"Constant" means anything that does not depend on xx, however it is written. The numbers π2\pi^2, e3e^3, ln⁡5\ln 5 and 17\sqrt{17} are all constants, so ddx(π2)=0\frac{d}{dx}\left(\pi^2\right) = 0. A letter that names a fixed parameter, such as kk in y=x2+ky = x^2 + k, is a constant too.

A constant multiplier passes through

If g(x)=k f(x)g(x) = k\,f(x) for a constant kk, every output of ff is multiplied by kk, so every rise on the graph is multiplied by kk while every run stays the same. Every slope is therefore multiplied by kk. In the limit definition, the kk factors out of the difference quotient,

k f(x+h)−k f(x)h=k⋅f(x+h)−f(x)h,\frac{k\,f(x+h) - k\,f(x)}{h} = k\cdot\frac{f(x+h) - f(x)}{h},

and the limit of a constant times something is the constant times the limit (the limit laws from topic 1.5). So g′(x)=k f′(x)g'(x) = k\,f'(x), provided ff is differentiable at xx.

A negative kk flips the graph upside down, and the slopes flip sign with it: if ff is rising with slope 4 at some point, −3f-3f is falling with slope −12-12 there.

Compare the two rules so far. In 5x45x^4 the 5 multiplies a function of xx, so it stays: ddx(5x4)=5⋅4x3=20x3\frac{d}{dx}\left(5x^4\right) = 5\cdot 4x^3 = 20x^3. In x4+5x^4 + 5 the 5 stands alone, so it disappears: ddx(x4+5)=4x3\frac{d}{dx}\left(x^4 + 5\right) = 4x^3. Adding a constant shifts the graph up or down, which changes no slope; multiplying by a constant stretches the graph, which changes every slope.

Sums and differences go term by term

If s(x)=f(x)+g(x)s(x) = f(x) + g(x), then a change in xx changes ss by the change in ff plus the change in gg. Dividing by hh splits the difference quotient into two:

s(x+h)−s(x)h=f(x+h)−f(x)h+g(x+h)−g(x)h.\frac{s(x+h) - s(x)}{h} = \frac{f(x+h) - f(x)}{h} + \frac{g(x+h) - g(x)}{h}.

When ff and gg are both differentiable at xx, each fraction on the right has a limit, and the limit of a sum is the sum of the limits. The same argument with a minus sign handles differences.

The hypothesis matters in one direction only. If both pieces are differentiable, so is the sum. If one piece is not, the rule tells you nothing, and you have to look at the whole function. For example, ∣x∣|x| is not differentiable at 0, yet ∣x∣+(−∣x∣)|x| + (-|x|) is the constant function 0, which is differentiable everywhere.

These rules also handle a function you know only by name. If f′(2)=3f'(2) = 3 and g′(2)=−1g'(2) = -1, then the derivative of 4f(x)−g(x)4f(x) - g(x) at x=2x = 2 is 4(3)−(−1)=134(3) - (-1) = 13, with no formula for ff or gg needed.

Polynomials and other sums of powers

The power rule from topic 2.5 says ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1} for any constant exponent nn. A polynomial is a sum of constant multiples of powers, so the three rules let you apply the power rule one term at a time:

ddx(4x3−7x+2)=4⋅3x2−7⋅1+0=12x2−7.\frac{d}{dx}\left(4x^3 - 7x + 2\right) = 4\cdot 3x^2 - 7\cdot 1 + 0 = 12x^2 - 7.

Each term drops one degree, the linear term −7x-7x becomes the constant −7-7, and the constant term vanishes. So for n≥1n \ge 1, the derivative of a polynomial of degree nn is a polynomial of degree n−1n - 1. A constant polynomial (degree 0) is the exception: its derivative is 0 everywhere.

The rules in this topic apply only to sums, differences and constant multiples. They do not cover a product or quotient of two functions of xx; topics 2.8 and 2.9 handle those. Many products and quotients can be rewritten as sums of powers first, though, and that rewrite is often the fastest route:

GivenRewritten as a sum of powersDerivative
(2x−3)2(2x - 3)^24x2−12x+94x^2 - 12x + 98x−128x - 12
x3−4xx\dfrac{x^3 - 4x}{x}x2−4x^2 - 4 (for x≠0x \neq 0)2x2x (for x≠0x \neq 0)
3x2\dfrac{3}{x^2}3x−23x^{-2}−6x−3=−6x3-6x^{-3} = -\dfrac{6}{x^3}
14x3\dfrac{1}{4x^3}14x−3\dfrac{1}{4}x^{-3}−34x−4-\dfrac{3}{4}x^{-4}
x (x+6)\sqrt{x}\,(x + 6)x3/2+6x1/2x^{3/2} + 6x^{1/2}32x1/2+3x−1/2\dfrac{3}{2}x^{1/2} + 3x^{-1/2}

Two details in that table matter. In the second row, the original function is undefined at x=0x = 0 even after you cancel, so its derivative is undefined there too. In the fourth row, the 4 sits in the denominator with x3x^3, so the coefficient is 14\frac{1}{4}, not 4.

Worked examples

Worked exampleExample 1: horizontal and parallel tangent lines

Let f(x)=x3−6x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1.

(a) Find all xx at which the line tangent to the graph of ff is horizontal.

(b) Find all xx at which the line tangent to the graph of ff is parallel to the line y=9x−4y = 9x - 4.

Show solution

Both parts are questions about slope, so start from the derivative. Term by term,

f′(x)=3x2−12x+9.f'(x) = 3x^2 - 12x + 9.

(a) A horizontal tangent line has slope 0. Set f′(x)=0f'(x) = 0 and factor:

3x2−12x+9=3(x2−4x+3)=3(x−1)(x−3)=0,3x^2 - 12x + 9 = 3\left(x^2 - 4x + 3\right) = 3(x - 1)(x - 3) = 0,

so the tangent line is horizontal at x=1x = 1 and at x=3x = 3.

(b) Parallel lines have equal slopes, and y=9x−4y = 9x - 4 has slope 9. Set f′(x)=9f'(x) = 9:

3x2−12x+9=9⟹3x(x−4)=0,3x^2 - 12x + 9 = 9 \quad\Longrightarrow\quad 3x(x - 4) = 0,

so x=0x = 0 or x=4x = 4. The constant −4-4 in the given line plays no part; only its slope does.

Worked exampleExample 2: rewrite, then differentiate

Let g(x)=x4−6x2xg(x) = \dfrac{x^4 - 6\sqrt{x}}{2x} for x>0x > 0. Find g′(x)g'(x) and g′(4)g'(4).

Show solution

The rules in this topic don't apply to a quotient, but dividing each term of the numerator by 2x2x turns gg into a sum of powers:

g(x)=x42x−6x1/22x=12x3−3x−1/2.g(x) = \frac{x^4}{2x} - \frac{6x^{1/2}}{2x} = \frac{1}{2}x^3 - 3x^{-1/2}.

Now apply the power rule to each term and keep each coefficient:

g′(x)=32x2−3(−12)x−3/2=32x2+32x−3/2.g'(x) = \frac{3}{2}x^2 - 3\left(-\frac{1}{2}\right)x^{-3/2} = \frac{3}{2}x^2 + \frac{3}{2}x^{-3/2}.

At x=4x = 4, 4−3/2=184^{-3/2} = \frac{1}{8}, so

g′(4)=32(16)+32⋅18=24+316=38716.g'(4) = \frac{3}{2}(16) + \frac{3}{2}\cdot\frac{1}{8} = 24 + \frac{3}{16} = \frac{387}{16}.

Rewriting before differentiating is the move whenever a quotient has a single power in its denominator: it avoids the quotient rule and leaves less algebra to get wrong.

Worked exampleExample 3: functions known only by values (no calculator)

The functions ff and gg are differentiable, with f(1)=2f(1) = 2, f′(1)=−3f'(1) = -3, g(1)=5g(1) = 5 and g′(1)=4g'(1) = 4. Let h(x)=2f(x)−3g(x)+x4h(x) = 2f(x) - 3g(x) + x^4. Write an equation for the line tangent to the graph of hh at x=1x = 1.

Show solution

A tangent line needs a point and a slope.

The point: h(1)=2f(1)−3g(1)+14=2(2)−3(5)+1=−10h(1) = 2f(1) - 3g(1) + 1^4 = 2(2) - 3(5) + 1 = -10.

The slope: by the sum, difference and constant multiple rules, h′(x)=2f′(x)−3g′(x)+4x3h'(x) = 2f'(x) - 3g'(x) + 4x^3, so

h′(1)=2(−3)−3(4)+4=−14.h'(1) = 2(-3) - 3(4) + 4 = -14.

The tangent line is y=−10−14(x−1)y = -10 - 14(x - 1).

The values f(1)f(1) and g(1)g(1) go into the point and the derivatives f′(1)f'(1) and g′(1)g'(1) go into the slope. Swapping them would give the point 2(−3)−3(4)+1=−172(-3) - 3(4) + 1 = -17 and the slope 2(2)−3(5)+4=−72(2) - 3(5) + 4 = -7, a different line entirely.

Worked exampleExample 4: an exam-style free-response part

Let f(x)=ax3+bx2−4f(x) = ax^3 + bx^2 - 4, where aa and bb are constants. The line y=5x−6y = 5x - 6 is tangent to the graph of ff at x=2x = 2. Find aa and bb.

Show solution

"Tangent at x=2x = 2" gives two facts: the curve and the line meet at x=2x = 2, and they have the same slope there.

The line passes through (2,5⋅2−6)=(2,4)(2, 5\cdot 2 - 6) = (2, 4), so f(2)=4f(2) = 4:

8a+4b−4=4⟹2a+b=2.8a + 4b - 4 = 4 \quad\Longrightarrow\quad 2a + b = 2.

The line has slope 5. Since f′(x)=3ax2+2bxf'(x) = 3ax^2 + 2bx (the constants aa and bb are multipliers, and −4-4 vanishes), f′(2)=5f'(2) = 5 gives

12a+4b=5.12a + 4b = 5.

Subtract 4 times the first equation, 8a+4b=88a + 4b = 8, from the second: 4a=−34a = -3, so a=−34a = -\dfrac{3}{4}. Then b=2−2a=72b = 2 - 2a = \dfrac{7}{2}.

Check both conditions: f(2)=−34(8)+72(4)−4=−6+14−4=4f(2) = -\frac{3}{4}(8) + \frac{7}{2}(4) - 4 = -6 + 14 - 4 = 4, and f′(2)=12(−34)+4(72)=−9+14=5f'(2) = 12\left(-\frac{3}{4}\right) + 4\left(\frac{7}{2}\right) = -9 + 14 = 5. Using only one of the two conditions leaves infinitely many pairs (a,b)(a, b), which is why the problem needs both.

Where students lose points

On the exam

These rules show up inside tangent-line problems, motion problems and optimization. On the no-calculator section you are expected to differentiate polynomials and sums of powers by hand. When a function is given only through values of ff and f′f', the constant multiple and sum rules are how you find the derivative of a combination such as 3f(x)−g(x)3f(x) - g(x).

Check yourself

  1. Question 1

    If f(x)=3x4−2x+π2f(x) = 3x^4 - \dfrac{2}{x} + \pi^2, then f′(x)=f'(x) =

    1. A

      12x3−2x−212x^3 - 2x^{-2}

    2. B

      12x3+2x−212x^3 + 2x^{-2}

    3. C

      12x3+2x−2+2π12x^3 + 2x^{-2} + 2\pi

    4. D

      12x3+2x−2+π212x^3 + 2x^{-2} + \pi^2

    Show answer

    B

    Rewrite −2x-\dfrac{2}{x} as −2x−1-2x^{-1}; its derivative is −2(−1)x−2=2x−2-2(-1)x^{-2} = 2x^{-2}, and π2\pi^2 is a constant with derivative 0. Choice A loses the sign from the exponent −1-1, and choice C differentiates the number π2\pi^2 as if π\pi were the variable.

  2. Question 2

    Let f(x)=2x3+5f(x) = 2x^3 + 5 and g(x)=2x3−8g(x) = 2x^3 - 8. How are f′f' and g′g' related, and what does that say about the two graphs?

    Show answer

    f′(x)=g′(x)=6x2f'(x) = g'(x) = 6x^2 for every xx. The graphs are vertical shifts of each other (by 13 units), so at every xx they have the same slope.

    The functions differ only by the constant 1313, whose derivative is 0. Shifting a graph up or down moves every point the same distance and changes no slope.

  3. Question 3

    The function ff is differentiable with f(2)=1f(2) = 1 and f′(2)=4f'(2) = 4. What is the slope of the graph of y=5−3f(x)y = 5 - 3f(x) at x=2x = 2?

    1. A

      −7-7

    2. B

      1212

    3. C

      −12-12

    4. D

      22

    Show answer

    C

    The derivative is 0−3f′(x)0 - 3f'(x), so the slope at x=2x = 2 is −3(4)=−12-3(4) = -12. Choice A keeps the constant 5, and choice D uses f(2)f(2) (the height, giving 5−3=25 - 3 = 2) instead of f′(2)f'(2) (the slope).

  4. Question 4

    Using only the rules of this topic, find ddx[(x+3)(x−3)]\dfrac{d}{dx}\big[(x+3)(x-3)\big]. Why is the answer not (1)(1)=1(1)(1) = 1?

    Show answer

    2x2x. Expand first: (x+3)(x−3)=x2−9(x+3)(x-3) = x^2 - 9, whose derivative is 2x2x.

    The sum and constant multiple rules split sums, not products. Multiplying the two derivatives 1⋅11 \cdot 1 would give a constant slope, but x2−9x^2 - 9 is a parabola whose slope changes with xx.

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