Grade by Grade
A. Rate of Change · Limits and Rates of Change

The Derivative at a Point: First Principles

A1.5 A1.6About 75 minutes
Learning goals
  • I can define the derivative of f at x = a as the limit of the difference quotient, and use the notation f′(a).
  • I can find f′(a) from first principles for polynomial, rational and radical functions.
  • I can use f′(a) to write the equation of a tangent line.
  • I can interpret f′(a) as an instantaneous rate of change in context, with units (velocity, marginal cost).

Words to know

derivative at a point
f′(a) = lim (h → 0) (f(a + h) − f(a))/h, if this limit exists.
first principles
Finding a derivative directly from its limit definition, without shortcut rules.
differentiable at a
A function is differentiable at a if f′(a) exists (the limit is a real number).
f′(a)
Read "f prime of a". The slope of the tangent at (a, f(a)) and the instantaneous rate of change of f at a.
Leibniz notation
dy/dx, the derivative written as a quotient of tiny changes; dy/dx at x = a means f′(a).
marginal cost
The instantaneous rate of change of total cost with respect to the number of items; close to the cost of making one more item.

It finally gets a name

The limit you have been calculating for several lessons is the most important limit in mathematics, so it gets its own name and notation.

The derivative of f at x = a is

f′(a) = lim (h → 0) (f(a + h) − f(a))/h

provided the limit exists. If it does, f is differentiable at a. The equivalent form f′(a) = lim (x → a) (f(x) − f(a))/(x − a) is sometimes easier.

f′(a) means three things at once:

  • Numerically: the limit of average rates over shrinking intervals.
  • Graphically: the slope of the tangent to y = f(x) at (a, f(a)).
  • In context: the instantaneous rate of change, in units of output per unit of input.

Calculating a derivative straight from this definition is called working from first principles. The sibling unit Derivatives and Differentiation Rules develops shortcuts, but they all come from this definition.

A reliable first-principles routine

  1. Write f(a) and f(a + h). Use brackets: f(a + h) means replace every x by (a + h).
  2. Subtract: f(a + h) − f(a). Every term without h should cancel. If not, look for an error.
  3. Simplify the quotient so that h is no longer in the denominator alone: expand, combine fractions, or rationalize.
  4. Let h → 0.
  5. State the answer with meaning: a slope, a rate, units.
Worked example

Example 1: A polynomial

For f(x) = 3x² − x, find f′(2) from first principles and the equation of the tangent at x = 2.

  1. f(2) = 12 − 2 = 10.
  2. f(2 + h) = 3(4 + 4h + h²) − (2 + h) = 12 + 12h + 3h² − 2 − h = 10 + 11h + 3h².
  3. f(2 + h) − f(2) = 11h + 3h², so the quotient is 11 + 3h.
  4. f′(2) = lim (h → 0) (11 + 3h) = 11.
  5. Tangent: y − 10 = 11(x − 2), so y = 11x − 12.

f′(2) = 11; tangent y = 11x − 12.

Worked example

Example 2: A rational function

For f(x) = 2/(x + 1), find f′(1) from first principles and the tangent at x = 1.

  1. f(1) = 2/2 = 1 and f(1 + h) = 2/(2 + h).
  2. f(1 + h) − f(1) = 2/(2 + h) − 1 = (2 − (2 + h))/(2 + h) = −h/(2 + h).
  3. Divide by h: −1/(2 + h).
  4. f′(1) = lim (h → 0) −1/(2 + h) = −1/2.
  5. Tangent: y − 1 = −½(x − 1), so y = −x/2 + 3/2.

f′(1) = −1/2; tangent y = −x/2 + 3/2.

012345123xy(1, 1)tangenty = 2/(x + 1)f′(1) = lim [f(1 + h) − f(1)]/hh→0= lim −1/(2 + h) = −1/2Tangent at (1, 1):y = −x/2 + 3/2It meets the x-axis at x = 3.
The graph of y = 2/(x + 1) and its tangent at (1, 1), which has slope f′(1) = −1/2.
Worked example

Example 3: A radical function

For f(x) = √(2x + 1), find f′(4).

  1. f(4) = √9 = 3 and f(4 + h) = √(2(4 + h) + 1) = √(9 + 2h).
  2. Quotient: (√(9 + 2h) − 3)/h. Multiply top and bottom by √(9 + 2h) + 3.
  3. Top: (9 + 2h) − 9 = 2h. So the quotient is 2h/(h(√(9 + 2h) + 3)) = 2/(√(9 + 2h) + 3).
  4. Let h → 0: 2/(3 + 3) = 1/3.

f′(4) = 1/3

Worked example

Example 4: Marginal cost

A small Canadian clothing company finds that the cost of making x hoodies in a run is C(x) = 500 + 12x − 0.01x² dollars (0 ≤ x ≤ 300). Find C′(100) and interpret it. Compare with the actual cost of the 101st hoodie.

  1. C(100) = 500 + 1200 − 100 = 1600.
  2. C(100 + h) = 500 + 12(100 + h) − 0.01(100 + h)² = 500 + 1200 + 12h − 0.01(10 000 + 200h + h²) = 1600 + 10h − 0.01h².
  3. Quotient: (10h − 0.01h²)/h = 10 − 0.01h. So C′(100) = 10.
  4. Actual cost of the 101st hoodie: C(101) − C(100) = 10(1) − 0.01(1)² = 9.99.

C′(100) = $10 per hoodie. When 100 hoodies are being made, the cost is rising at about $10 per extra hoodie, almost exactly the true extra cost of $9.99. Economists call C′(x) the marginal cost.

050100150200250300050010001500200025003000x (hoodies)C ($)(100, 1600)dashed: tangent at x = 100,slope 10C(x) = 500 + 12x − 0.01x²C′(100) = $10 per hoodie(marginal cost)Real cost of hoodie 101:C(101) − C(100) = $9.99The tangent slope is anexcellent estimate.
Total cost of a run of hoodies. The tangent at x = 100 has slope C′(100) = 10, the marginal cost. The cost curve bends downward because each extra hoodie costs a little less in a larger run.
Worked example

Example 5: Using the x → a form

Find f′(−1) for f(x) = x³ using f′(a) = lim (x → a) (f(x) − f(a))/(x − a).

  1. f(−1) = −1, so the quotient is (x³ + 1)/(x + 1).
  2. Sum of cubes: x³ + 1 = (x + 1)(x² − x + 1).
  3. Cancel x + 1: x² − x + 1.
  4. Let x → −1: 1 + 1 + 1 = 3.

f′(−1) = 3. (Matches 3a² from the last lesson: 3(−1)² = 3.)

Just for fun

Gottfried Wilhelm Leibniz published his calculus in 1684, and his notation dy/dx, a quotient of tiny differences, is still the one on your formula sheet more than 340 years later. Very few inventions from the 1680s are still in daily use. (Newton developed calculus at about the same time, and the two had a famously grumpy argument about who did it first.)

Hands-on activity

Marginal cost of a batch of cookies

You need: a recipe for cookies (or any batch recipe), current grocery prices (a flyer or website), a calculator, and paper.

  1. Work out the total cost of the ingredients for one batch, two batches and four batches. Include a fixed cost that does not depend on batch size, such as oven energy for one bake (estimate about $0.50) or a reusable baking sheet (spread its price over many uses).
  2. Suppose bulk buying saves you money, so the cost of x batches is modelled by something like C(x) = F + px − qx², where F is the fixed cost, p the price per batch, and q a small bulk discount. Choose values that fit your numbers.
  3. Find C′(2) from first principles. Interpret it: what does one more batch cost, when you are already making two?
  4. Compare C′(2) with C(3) − C(2).

What to look for: with any model of the form C(x) = F + px − qx², first principles give C′(a) = p − 2qa. The marginal cost is a little less than p and falls as you make more batches. C′(2) and C(3) − C(2) differ by exactly q, a small amount, so the derivative is a good estimate of the cost of one more batch. (Ask a grown-up before using the oven.)

Your turn

1

Find f′(3) for f(x) = x² + 2x from first principles.

2

Find f′(−1) for f(x) = 5 − x².

3

Find f′(3) for f(x) = 1/x.

4

Find f′(9) for f(x) = √x.

5

Write the tangent to f(x) = x² + 2x at x = 3.

6

A ball rolls down a long ramp. Its distance is s(t) = 0.5t² metres after t seconds. Find s′(4) and interpret it.

7

T(t) is the temperature in °C, t hours after midnight, and T′(5) = −3. Interpret this.

More practice: warm-up

1

How do you read f′(2) aloud, and what does it mean?

2

What does f′(a) represent on the graph of f?

3

For f(x) = 7x − 3, what is f′(a) for any a?

4

For a constant function f(x) = c, what is f′(a)?

5

f′(4) = 0. What does the tangent at x = 4 look like?

More practice: on your own

1

Find f′(1) for f(x) = 2x³.

2

Find f′(5) for f(x) = 3/(x − 2).

3

Find f′(4) for f(x) = √(x + 5).

4

For C(x) = 500 + 12x − 0.01x², find C′(200) from first principles and interpret it.

5

Find the tangent to y = 1/x at x = 1/2.

Challenge

1

Find f′(a) for f(x) = 1/x² from first principles.

2

Find the point on y = √x where the tangent is parallel to y = x/4.

3

Show that the tangent to y = x² at (a, a²) has y-intercept −a².

Check your understanding

Quiz

1.

Which expression is the definition of f′(a)?

2.

For f(x) = 3x² − x, the difference quotient at a = 2 simplifies to 11 + 3h. What is f′(2)?

3.

What is f′(1) for f(x) = 2/(x + 1)?

4.

C(x) is the cost in dollars of making x items, and C′(150) = 7. What does this mean?

5.

The tangent to y = f(x) at (2, 5) has slope f′(2) = −3. What is its equation?

6.

What is f′(4) for f(x) = √x?

For parents and teachers

Big idea (A1.5, A1.6): the derivative f′(a) is defined as the limit of the difference quotient. It is the slope of the tangent and the instantaneous rate of change at a. Working from first principles means using this definition directly, for polynomial, rational and radical functions.

How to help: Ask your teen to explain the five-step routine in the tip box, then to interpret one answer in words with units ("At 100 hoodies, each extra hoodie costs about $10").

Common difficulties:

  • Writing f(a + h) as f(a) + h, or forgetting brackets: f(2 + h) for 3x² − x is 3(2 + h)² − (2 + h).
  • Terms without h that do not cancel in f(a + h) − f(a): this always signals an arithmetic error.
  • Forgetting to multiply both top and bottom by the conjugate.
  • Finding the slope but writing the tangent through the wrong point; use (a, f(a)).