Mathematics Advanced • Year 11 • Module 3 • Lesson 3

The Derivative as the Gradient of a Tangent

Build procedural fluency in differentiating from first principles — expand, simplify, factor h, cancel, then take the limit.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Write the limit definition of the derivative (first principles):

f ′(x) = lim_{h→__} ( ____________________ ) / ____ .

Q1.2 Two notations for the derivative are:

Lagrange: ____________ and Leibniz: ____________ . Both represent the same quantity.

Q1.3 The gradient of the tangent to y = f(x) at the point where x = a is given by ________ (write a symbol). The gradient of the normal at the same point is ________ .

Stuck? Revisit lesson § Key Terms.

2. Worked example — f ′(x) for f(x) = x² from first principles

Follow each line of algebra. Every step has a reason on the right.

Problem. Find f ′(x) for f(x) = x² from first principles.

Step 1 — Form f(x + h).

f(x + h) = (x + h)² = x² + 2xh + h²

Reason: replace every x with (x + h) and expand fully.

Step 2 — Form the difference quotient and simplify the numerator.

( f(x + h) − f(x) ) / h = ( x² + 2xh + h² − x² ) / h = ( 2xh + h² ) / h

Reason: the x² terms cancel — only the "new" parts survive.

Step 3 — Factor h out of the numerator and cancel.

( h(2x + h) ) / h = 2x + h (for h ≠ 0)

Reason: cancelling h is valid because we are taking a limit, not setting h = 0.

Step 4 — Take the limit as h → 0.

f ′(x) = lim_h→0 (2x + h) = 2x

Conclusion. The derivative of x² is f ′(x) = 2x.

3. Faded example — fill in the missing steps

Find f ′(x) for f(x) = 3x + 1 from first principles. Fill in each blank. 3 marks

Step 1 — Form f(x + h):

f(x + h) = 3( ________ ) + 1 = ________________

Step 2 — Difference quotient:

( f(x + h) − f(x) ) / h = ( ________________ − (3x + 1) ) / h = ________ / h

Step 3 — Cancel h:

( ________ ) / h = ________ (for h ≠ 0)

Step 4 — Take the limit:

f ′(x) = lim_h→0 ( ________ ) = ________ .

Conclusion. f ′(x) = ________ . (Notice: for a linear function, the derivative is just the slope.)

Stuck? Revisit lesson § Worked Example 2.

4. Graduated practice — differentiate and use the derivative

For each function, follow the four-step first-principles process. For questions about tangents/normals, work from the derivative you find.

Foundation — first principles, simple terms (4 questions)

Show every step of the four-step process.

4.1 Differentiate f(x) = 5 from first principles. 1 mark

4.2 Differentiate f(x) = 2x + 3 from first principles. 1 mark

4.3 Differentiate f(x) = x² from first principles. 1 mark

4.4 Differentiate f(x) = 3x² from first principles. 1 mark

Standard — typical HSC difficulty (6 questions)

4.5 Differentiate f(x) = x² + 3x from first principles. 2 marks

4.6 Differentiate f(x) = x² − 5 from first principles. 2 marks

4.7 Differentiate f(x) = x³ from first principles. (Hint: (x + h)³ = x³ + 3x²h + 3xh² + h³.) 2 marks

4.8 Using the result f ′(x) = 2x for f(x) = x², find the gradient of the tangent to y = x² at x = 4. 2 marks

4.9 Find the equation of the tangent to y = x² at the point (1, 1). 2 marks

4.10 Find the equation of the normal to y = x² at the point (2, 4). 2 marks

Extension — combine concepts (2 questions)

4.11 Differentiate f(x) = 2x² + 3x − 1 from first principles. Show every step. 3 marks

4.12 A tangent to y = x² at the point P(a, a²) has gradient 6. Find the value of a, and hence write the equation of this tangent. 3 marks

Stuck on 4.12? f ′(x) = 2x, so set 2a = 6.

5. Self-check the easy 3

Tick the first three once you have checked your method works.

For 4.1 (constant function) I got f ′(x) = 0 — the difference (5 − 5) is always 0.

For 4.2 the difference quotient simplified to a constant (2) before the limit was taken — linear functions have constant derivative.

For 4.3 I expanded (x + h)² as x² + 2xh + h² (three terms, not two) and arrived at 2x.

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Answers — Do not peek before attempting

Q1.1 — Limit definition

f ′(x) = lim_h→0 ( f(x + h) − f(x) ) / h.

Q1.2 — Notation

Lagrange: f ′(x). Leibniz: dy/dx.

Q1.3 — Tangent and normal gradients at x = a

Tangent gradient = f ′(a). Normal gradient = −1 / f ′(a) (negative reciprocal).

Q3 — Faded example f(x) = 3x + 1

Step 1: f(x + h) = 3(x + h) + 1 = 3x + 3h + 1. Step 2: ( (3x + 3h + 1) − (3x + 1) ) / h = 3h / h. Step 3: 3h / h = 3 (for h ≠ 0). Step 4: lim_h→0 3 = 3. Conclusion: f ′(x) = 3.

Q4.1 — f(x) = 5

f(x + h) = 5. (5 − 5)/h = 0/h = 0 (for h ≠ 0). lim_h→0 0 = 0. The derivative of any constant is 0.

Q4.2 — f(x) = 2x + 3

f(x + h) = 2(x + h) + 3 = 2x + 2h + 3. ( (2x + 2h + 3) − (2x + 3) ) / h = 2h / h = 2. lim_h→0 2 = 2.

Q4.3 — f(x) = x²

f(x + h) = x² + 2xh + h². (2xh + h²) / h = 2x + h. lim_h→0 (2x + h) = 2x.

Q4.4 — f(x) = 3x²

f(x + h) = 3(x + h)² = 3x² + 6xh + 3h². (6xh + 3h²) / h = 6x + 3h. lim_h→0 (6x + 3h) = 6x. (Consistent with constant-multiple: 3 × derivative of x² = 3 × 2x = 6x.)

Q4.5 — f(x) = x² + 3x

f(x + h) = (x + h)² + 3(x + h) = x² + 2xh + h² + 3x + 3h. Subtract f(x) = x² + 3x: leaves 2xh + h² + 3h. Divide by h: 2x + h + 3. lim_h→0 (2x + h + 3) = 2x + 3.

Q4.6 — f(x) = x² − 5

f(x + h) = (x + h)² − 5 = x² + 2xh + h² − 5. Subtract f(x) = x² − 5: leaves 2xh + h². Divide by h: 2x + h. lim_h→0 = 2x. (Adding a constant does not change the derivative — the constant's derivative is 0.)

Q4.7 — f(x) = x³

f(x + h) = x³ + 3x²h + 3xh² + h³. Subtract x³: leaves 3x²h + 3xh² + h³. Divide by h: 3x² + 3xh + h². lim_h→0 = 3x².

Q4.8 — Tangent gradient to y = x² at x = 4

f ′(x) = 2x, so f ′(4) = 2(4) = 8.

Q4.9 — Equation of tangent to y = x² at (1, 1)

Gradient = f ′(1) = 2. Point-gradient: y − 1 = 2(x − 1) ⇒ y = 2x − 1.

Q4.10 — Equation of normal to y = x² at (2, 4)

Tangent gradient = f ′(2) = 4. Normal gradient = −1/4. y − 4 = −1/4 · (x − 2) ⇒ y = −x/4 + 9/2 (or equivalently 4y + x = 18).

Q4.11 — f(x) = 2x² + 3x − 1

f(x + h) = 2(x + h)² + 3(x + h) − 1 = 2x² + 4xh + 2h² + 3x + 3h − 1. Subtract f(x) = 2x² + 3x − 1: leaves 4xh + 2h² + 3h. Divide by h: 4x + 2h + 3. lim_h→0 = 4x + 3.

Q4.12 — Tangent of gradient 6 to y = x²

f ′(x) = 2x. Set 2a = 6 ⇒ a = 3. Point P(3, 9). Tangent: y − 9 = 6(x − 3) ⇒ y = 6x − 9.