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Calculus I (Differential & intro Integral) Unit 2: Definition of the derivative

The derivative measures the instantaneous rate of change of a function, which is also the slope of its tangent line. It's defined as a limit of average rates of change.

Covers: Difference quotient, derivative as rate/slope, differentiability vs continuity, power/sum/product/quotient rules

Study guide

The derivative measures instantaneous rate of change. Start with the difference quotient [f(x + h) − f(x)]/h, which is the slope of a secant line, the average rate of change over a small interval. Taking the limit as h → 0 gives the derivative: f′(x) = lim h→0 [f(x + h) − f(x)]/h. Geometrically, f′(a) is the slope of the tangent line at x = a; in context it's a rate with units of output per input.

A function is differentiable at a point if this limit exists. Differentiability implies continuity, but not the other way: a continuous function can fail to be differentiable at a corner (like |x| at 0), a cusp, or a vertical tangent. Any discontinuity also prevents differentiability.

Rules make derivatives fast. The power rule gives d/dx xⁿ = nxⁿ⁻¹. The constant multiple and sum rules let you differentiate term by term. The product rule is (fg)′ = f′g + fg′, and the quotient rule is (f/g)′ = (f′g − fg′)/g². Memorize the basic derivatives: d/dx sin x = cos x, d/dx cos x = −sin x, d/dx eˣ = eˣ, and d/dx ln x = 1/x.

Use the derivative to write tangent lines with point-slope form: y − f(a) = f′(a)(x − a).

An equivalent definition is f′(a) = lim x→a [f(x) − f(a)]/(x − a). If you recognize a limit in either form as a derivative, you can evaluate it immediately: lim x→3 (ln x − ln 3)/(x − 3) is the derivative of ln x at 3, which is 1/3. When f is known only from a table, estimate f′(a) with the average rate of change over the smallest table interval that contains a, [f(b) − f(c)]/(b − c), and give units of output per unit of input.

Two curves have parallel tangent lines at x = c exactly when f′(c) = g′(c), and a graph has a horizontal tangent where f′(c) = 0. To find such points, set up the equation in the derivatives and solve it.

Key ideas

Formulas and rules

d/dx xⁿ = n·xⁿ⁻¹
(fg)′ = f′g + fg′
(f/g)′ = (f′g − fg′) / g²
d/dx sin x = cos x, d/dx cos x = −sin x
d/dx eˣ = eˣ, d/dx ln x = 1/x

Common mistakes

Words to know

Limit definition of the derivative
f′(x) = lim h→0 [f(x + h) − f(x)] / h
Power rule
d/dx xⁿ = n·xⁿ⁻¹
Product rule
(fg)′ = f′g + fg′
Quotient rule
(f/g)′ = (f′g − fg′) / g²
d/dx cos x
−sin x
d/dx ln x
1/x (for x > 0)
Differentiability vs. continuity
Differentiable implies continuous, but not the reverse (|x| at 0).
Where differentiability fails
Corners, cusps, vertical tangents and discontinuities.
Derivative
f′(x): the instantaneous rate of change and the slope of the tangent line.
Difference quotient
[f(x + h) − f(x)]/h: an average rate of change.
Tangent line equation
y − f(a) = f′(a)(x − a)
Constant multiple rule
(cf)′ = c·f′
Sum rule (derivatives)
(f + g)′ = f′ + g′
d/dx sin x
cos x
d/dx eˣ
eˣ

Sample practice questions

37 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

d/dx x⁵

  1. 5x⁴
  2. 4x⁵
  3. x⁴
  4. 5x⁵
Show the answer and steps

Answer: A) 5x⁴

  1. Power rule: d/dx xⁿ = n·xⁿ⁻¹.
  2. d/dx x⁵ = 5x⁴.
Question 2 (level 1)

Find d/dx (x⁵).

  1. x⁴
  2. 5x⁴
  3. 5x⁵
  4. 4x⁴
Show the answer and steps

Answer: B) 5x⁴

  1. Power rule: d/dx (axⁿ) = a·n·xⁿ⁻¹.
  2. 1·5·x⁴ = 5x⁴.
Question 3 (level 3)

d/dx (x² sin x)

  1. 2x sin x + x² cos x
  2. 2x sin x − x² cos x
  3. x² cos x
  4. 2x cos x
Show the answer and steps

Answer: A) 2x sin x + x² cos x

  1. Product rule: (fg)′ = f′g + fg′ with f = x², g = sin x.
  2. f′ = 2x, g′ = cos x.
  3. 2x sin x + x² cos x.