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Differentiation

2.1 The Derivative and the Tangent Line Problem

Focus on derivative from the definition, tangent slope, instantaneous velocity.

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Ideas in this lesson

01Derivative from the definitionWorked steps · five questions02Tangent slopeWorked steps · five questions03Instantaneous velocityWorked steps · five questions
Idea 1 of 3

Derivative from the definition

Simplify the difference quotient before taking h→0h \to 0. The derivative is a limit of secant slopes.

Worked example 1

Use the derivative definition for f(x)=x2f(x) = x^{2}. Find f′(1)f'(1).
  1. (x+h)2−x2h=2x+h\dfrac{(x + h)^{2} - x^{2}}{h} = 2x + h
  2. f′(x)=2xf'(x) = 2x
  3. f′(1)=2(1)=2f'(1) = 2(1) = 2

Answer: 22

Worked example 2

Use the derivative definition for f(x)=x2+3xf(x) = x^{2} + 3x. Find f′(2)f'(2).
  1. (x+h)2−x2+3hh=2x+h+3\dfrac{(x + h)^{2} - x^{2} + 3h}{h} = 2x + h + 3
  2. f′(x)=2x+3f'(x) = 2x + 3
  3. f′(2)=2(2)+3=7f'(2) = 2(2) + 3 = 7

Answer: 77

Idea 2 of 3

Tangent slope

The derivative at an input is the tangent slope there, when the derivative exists.

Worked example 1

For f(x)=x2f(x) = x^{2}, find the secant slope from x=2x = 2 to x=3x = 3.
  1. m=(x+h)2−x2h=2x+hm = \dfrac{(x + h)^{2} - x^{2}}{h} = 2x + h
  2. m=2(2)+1=5m = 2(2) + 1 = 5
  3. f′(2)=lim⁡h→0(2(2)+h)=4\displaystyle f'(2) = \lim\limits_{h\to 0} (2(2) + h)= 4

Answer: 55

Worked example 2

For f(x)=x2f(x) = x^{2}, find the secant slope from x=3x = 3 to x=3.1x = 3.1.
  1. m=(x+h)2−x2h=2x+hm = \dfrac{(x + h)^{2} - x^{2}}{h} = 2x + h
  2. m=2(3)+0.1=6110≈6.1m = 2(3) + 0.1 = \dfrac{61}{10} \approx 6.1
  3. f′(3)=lim⁡h→0(2(3)+h)=6\displaystyle f'(3) = \lim\limits_{h\to 0} (2(3) + h)= 6

Answer: 6110≈6.1\dfrac{61}{10} \approx 6.1

Idea 3 of 3

Instantaneous velocity

Differentiate position to obtain velocity; average velocity still uses a secant quotient.

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, §2.1,p2.1, p. 9696. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.