Unit 2 · Topic 2.6 Beta

Constant, Sum, Difference and Constant Multiple Rules

4 min read · freeNot practiced

The power rule handles one power of xx at a time. Real functions are built from several: 4x3−5x+24x^3 - 5x + 2, or 3f(x)−2g(x)3f(x) - 2g(x). This page gives the rules that take derivatives through constants, constant factors, sums and differences. With them and the power rule, every polynomial can be differentiated in one pass.

The constant rule

A constant function, f(x)=cf(x) = c, has a level graph. A level line has slope 0 everywhere, so

ddx(c)=0.\frac{d}{dx}(c) = 0.

The same idea explains what happens when you add a constant to a function. Adding 3 lifts the whole graph 3 units without tilting it, so every tangent line keeps its slope.

The parabola y = f(x) = 0.5x squared minus x, and the same curve shifted up 3 units, y = f(x) + 3. At x = 2.5 each curve has a dashed tangent line; the two tangent lines are parallel, both with slope 1.5. Adding a constant changes heights but no slopes.
Figure 1. ff and f+3f + 3 have parallel tangent lines at every xx. LevlPrep original diagram.

Watch for constants that look like variables: π2\pi^2, e3e^3 and 5\sqrt{5} are numbers, and their derivatives are 0.

The constant multiple rule

Multiply a function by 4 and every height is multiplied by 4. Over the same run, every rise is 4 times as big, so every slope is 4 times as big:

ddx(kf(x))=kf′(x).\frac{d}{dx}\left( kf(x) \right) = kf'(x).

A constant factor rides along. So ddx(7x3)\frac{d}{dx}\left( 7x^3 \right)=7⋅3x2= 7 \cdot 3x^2=21x2= 21x^2, and ddx(9x)=9\frac{d}{dx}(9x) = 9.

Keep the two constant rules apart. A constant term (added) disappears; a constant factor (multiplied) stays:

FunctionThe constant is aDerivative
x2+7x^2 + 7term2x2x
7x27x^2factor14x14x
77the whole function00
7x7xfactor77

The sum and difference rules

If ff and gg change together, the change in f+gf + g is the change in ff plus the change in gg. Divide by the same hh and take the limit, and the rates add:

ddx(f(x)+g(x))=f′(x)+g′(x),ddx(f(x)−g(x))=f′(x)−g′(x).\frac{d}{dx}\left( f(x) + g(x) \right) = f'(x) + g'(x), \qquad \frac{d}{dx}\left( f(x) - g(x) \right) = f'(x) - g'(x).

Two pumps adding 30 and 45 liters a minute fill a pool at 75 liters a minute. If one drains at 20 while the other fills at 45, the level changes at 45−20=2545 - 20 = 25.

Polynomials, term by term

A polynomial is a sum of constant multiples of powers of xx, so the rules take it apart one term at a time.

Worked example 1. Differentiate f(x)=x3−6x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1 and find where its tangent line is level.

Term by term: ddx(x3)=3x2\frac{d}{dx}\left( x^3 \right) = 3x^2; ddx(−6x2)=−12x\frac{d}{dx}\left( -6x^2 \right) = -12x; ddx(9x)=9\frac{d}{dx}(9x) = 9; ddx(1)=0\frac{d}{dx}(1) = 0.

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

Level tangent line: 3x2−12x3x^2 - 12x+9+ 9=3(x2−4x+3)= 3(x^2 - 4x + 3)=3(x−1)(x−3)= 3(x - 1)(x - 3)=0= 0 at x=1x = 1 and x=3x = 3. On the graph these are the top of the hump and the bottom of the dip.

Worked example 2. Differentiate y=4x3−5x+2xy = 4x^3 - \frac{5}{x} + 2\sqrt{x}.

Rewrite every term as a constant times a power: yy=4x3−5x−1+2x1/2= 4x^3 - 5x^{-1} + 2x^{1/2}.

Differentiate: y′y'=12x2−5(−1)x−2= 12x^2 - 5(-1)x^{-2}+2⋅12x−1/2+ 2 \cdot \frac{1}{2}x^{-1/2}=12x2= 12x^2+5x−2+ 5x^{-2}+x−1/2+ x^{-1/2}.

Signs need care: −5-5 times −1-1 is +5+5.

Expand or divide first

The rules cover sums and constant multiples. They say nothing yet about products, quotients or powers of whole expressions, so turn those into sums first when the algebra is easy:

  • (x2+1)2=x4+2x2+1(x^2 + 1)^2 = x^4 + 2x^2 + 1, so its derivative is 4x3+4x4x^3 + 4x. Writing 2(x2+1)2(x^2 + 1) treats x2+1x^2 + 1 as if it were xx, which it is not.
  • x3−2xx=x2−2\frac{x^3 - 2x}{x} = x^2 - 2 for x≠0x \ne 0, so its derivative is 2x2x. Differentiating top and bottom separately gives 3x2−23x^2 - 2, which is wrong.

Rules for products and quotients that cannot be simplified come in topics 2.8 and 2.9.

Rules on tables of values

The rules work at a single input too. If h(x)=3f(x)−2g(x)h(x) = 3f(x) - 2g(x), then h′(2)=3f′(2)−2g′(2)h'(2) = 3f'(2) - 2g'(2). You need only the derivative values at x=2x = 2, not formulas.

Worked example 3. A table gives f(2)=3f(2) = 3, f′(2)=−1f'(2) = -1, g(2)=5g(2) = 5, g′(2)=4g'(2) = 4. Write the line tangent to h(x)=3f(x)−2g(x)h(x) = 3f(x) - 2g(x) at x=2x = 2.

Height: h(2)=3(3)−2(5)=−1h(2) = 3(3) - 2(5) = -1. Use the value columns.

Slope: h′(2)h'(2)=3(−1)−2(4)= 3(-1) - 2(4)=−11= -11. Use the derivative columns.

Line: y=−1−11(x−2)y = -1 - 11(x - 2). Mixing up the columns (values for the slope, or derivatives for the height) is the slip to avoid.

Common mistakes

  • Keeping a constant term: ddx(3x2+7)\frac{d}{dx}\left( 3x^2 + 7 \right) is 6x6x, not 6x+76x + 7.
  • Dropping a constant factor: ddx(6x2)\frac{d}{dx}\left( 6x^2 \right) is 12x12x, not 2x2x.
  • Losing a sign inside a difference: distribute the minus over the whole derivative of the second term.

Spot a mistake on this page?