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Differentiation

2.2 Basic Differentiation Rules and Rates of Change

Focus on power and constant-multiple rules, derivative of a constant, sum and difference rules, derivative of sine, derivative of cosine, derivative of tangent, velocity, acceleration, and speed.

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Ideas in this lesson

01Power and constant-multiple rulesWorked steps · five questions02Derivative of a constantWorked steps · five questions03Sum and difference rulesWorked steps · five questions04Derivative of sineWorked steps · five questions05Derivative of cosineWorked steps · five questions06Derivative of tangentWorked steps · five questions07Velocity, acceleration, and speedWorked steps · five questions
Idea 1 of 7

Power and constant-multiple rules

Differentiate term by term. A constant differentiates to zero; a constant factor stays outside the derivative.

Worked example 1

For f(x)=2x2−xf(x) = 2x^{2} - x, find f′(1)f'(1).
  1. f′(x)=4x−1f'(x) = 4x - 1
  2. f′(1)=3f'(1) = 3

Answer: 33

Worked example 2

For f(x)=3x3−xf(x) = 3x^{3} - x, find f′(2)f'(2).
  1. f′(x)=9x2−1f'(x) = 9x^{2} - 1
  2. f′(2)=35f'(2) = 35

Answer: 3535

Idea 2 of 7

Derivative of a constant

For a constant c,d(c)dx=0c, \frac{d(c)}{dx} = 0. A horizontal graph has zero slope.

Worked example 1

Let f(x)=2f(x) = 2. Find f′(1)f'(1).
  1. f(x)=2f(x) = 2
  2. f′(x)=0f'(x) = 0
  3. f′(1)=0f'(1) = 0

Answer: 00

Worked example 2

Let f(x)=6f(x) = 6. Find f′(2)f'(2).
  1. f(x)=6f(x) = 6
  2. f′(x)=0f'(x) = 0
  3. f′(2)=0f'(2) = 0

Answer: 00

Idea 3 of 7

Sum and difference rules

(f+g)′=f′+g′(f + g)' = f' + g' and (f−g)′=f′−g′(f - g)' = f' - g'. Differentiate each term separately.

Worked example 1

Let f(x)=2x2+3x−2f(x) = 2x^{2} + 3x - 2. Find f′(1)f'(1).
  1. f′(x)=2(2)x+3−0f'(x) = 2(2)x + 3 - 0
  2. f′(1)=4(1)+3f'(1) = 4(1) + 3
  3. f′(1)=7f'(1) = 7

Answer: 77

Worked example 2

Let f(x)=3x3+4x−3f(x) = 3x^{3} + 4x - 3. Find f′(2)f'(2).
  1. f′(x)=3(3)x2+4−0f'(x) = 3(3)x^{2} + 4 - 0
  2. f′(2)=9(2)2+4f'(2) = 9(2)^{2} + 4
  3. f′(2)=40f'(2) = 40

Answer: 4040

Idea 4 of 7

Derivative of sine

d(sin⁡x)dx=cos⁡x\frac{d(\sin x)}{dx} = \cos x, with xx measured in radians.

Worked example 1

Let f(x)=2sin⁡xf(x) = 2 \sin x. Find f′(0)f'(0).
  1. f′(x)=2cos⁡xf'(x) = 2 \cos x
  2. f′(0)=2cos⁡(0)f'(0) = 2 \cos (0)
  3. f′(0)=2f'(0) = 2

Answer: 22

Worked example 2

Let f(x)=3sin⁡xf(x) = 3 \sin x. Find f′(π2)f'(\dfrac{\pi }{2}).
  1. f′(x)=3cos⁡xf'(x) = 3 \cos x
  2. f′(π2)=3cos⁡(π2)f'(\dfrac{\pi }{2}) = 3 \cos (\dfrac{\pi }{2})
  3. f′(π2)=0f'(\dfrac{\pi }{2}) = 0

Answer: 00

Idea 5 of 7

Derivative of cosine

d(cos⁡x)dx=−sin⁡x\frac{d(\cos x)}{dx} = -\sin x. Keep the negative sign.

Worked example 1

Let f(x)=2cos⁡xf(x) = 2 \cos x. Find f′(0)f'(0).
  1. f′(x)=−2sin⁡xf'(x) = -2 \sin x
  2. f′(0)=−2sin⁡(0)f'(0) = -2 \sin (0)
  3. f′(0)=0f'(0) = 0

Answer: 00

Worked example 2

Let f(x)=3cos⁡xf(x) = 3 \cos x. Find f′(π2)f'(\dfrac{\pi }{2}).
  1. f′(x)=−3sin⁡xf'(x) = -3 \sin x
  2. f′(π2)=−3sin⁡(π2)f'(\dfrac{\pi }{2}) = -3 \sin (\dfrac{\pi }{2})
  3. f′(π2)=−3f'(\dfrac{\pi }{2}) = -3

Answer: −3-3

Idea 6 of 7

Derivative of tangent

d(tan⁡x)dx=sec⁡2x\frac{d(\tan x)}{dx} = \sec ^{2} x wherever cos⁡x≠0\cos x \ne 0.

Worked example 1

Let f(x)=2tan⁡xf(x) = 2 \tan x. Find f′(0)f'(0).
  1. f′(x)=2sec⁡2xf'(x) = 2 \sec ^{2} x
  2. f′(0)=2cos⁡2(0)f'(0) = \dfrac{2}{\cos ^{2}(0)}
  3. f′(0)=2f'(0) = 2

Answer: 22

Worked example 2

Let f(x)=3tan⁡xf(x) = 3 \tan x. Find f′(π4)f'(\dfrac{\pi }{4}).
  1. f′(x)=3sec⁡2xf'(x) = 3 \sec ^{2} x
  2. f′(π4)=3cos⁡2(π4)f'(\dfrac{\pi }{4}) = \dfrac{3}{\cos ^{2}(\dfrac{\pi }{4})}
  3. f′(π4)=6f'(\dfrac{\pi }{4}) = 6

Answer: 66

Idea 7 of 7

Velocity, acceleration, and speed

Velocity is position’ss first derivative; acceleration is its second derivative. Speed is the magnitude of velocity.

Worked example 1

A particle has position s(t)=t3−2ts(t) = t^{3} - 2t. Find its velocity at t=1t = 1.
  1. v(t)=s′(t)=3t2−2v(t) = s'(t) = 3t^{2} - 2
  2. a(t)=v′(t)=6ta(t) = v'(t) = 6t
  3. v(1)=3(1)2−2=1v(1) = 3(1)^{2} - 2 = 1

Answer: 11

Worked example 2

A particle has position s(t)=t3−3ts(t) = t^{3} - 3t. Find its acceleration at t=2t = 2.
  1. v(t)=s′(t)=3t2−3v(t) = s'(t) = 3t^{2} - 3
  2. a(t)=v′(t)=6ta(t) = v'(t) = 6t
  3. a(2)=6(2)=12a(2) = 6(2) = 12

Answer: 1212

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §2.2,p2.2, p. 106106. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.