Mathematics Advanced • Year 11 • Module 3 • Lesson 3
The Derivative as the Gradient of a Tangent
HSC-style writing on first principles — short answers plus an extended response deriving a tangent & normal at a point.
1. Short-answer questions
1.1 Find the derivative of f(x) = x² + 3x from first principles, showing every step. 3 marks Band 3
1.2 The curve y = x³ passes through the point (1, 1). Find the equation of the tangent to the curve at this point, using the fact (which you may quote) that d/dx (x³) = 3x². 3 marks Band 3-4
1.3 Explain in 2-3 sentences geometrically why the derivative of any linear function f(x) = mx + c is constant and equal to m. You may refer to a small sketch. 3 marks Band 4
Stuck on 1.3? "The tangent to a straight line at any point is the line itself." Build from there.2. Extended response
2.1 Consider the parabola y = x² − 4x + 5.
(a) Using first principles (the limit definition of the derivative), find dy/dx.
(b) Find the coordinates of the point on the curve where the tangent is parallel to the x-axis.
(c) Find the equation of the tangent to the curve at the point where x = 3.
(d) Find the equation of the normal to the curve at the same point (x = 3).
(e) Show that the tangent in (c) and the normal in (d) meet at the point (3, 2). 8 marks Band 5-6
Explicit marking criteria
Part (a) — 3 marks
• 1 mark — correctly forms f(x + h) = (x + h)² − 4(x + h) + 5 and expands.
• 1 mark — correctly forms the difference quotient and simplifies the numerator (cancels constant and linear x terms).
• 1 mark — factors h, cancels (with "for h ≠ 0"), takes the limit, and concludes dy/dx = 2x − 4.
Part (b) — 1 mark
• 1 mark — sets 2x − 4 = 0 ⇒ x = 2; computes y = 1; states the point (2, 1).
Part (c) — 1 mark
• 1 mark — finds gradient = 2; point (3, 2); tangent equation y = 2x − 4 (or equivalent).
Part (d) — 1 mark
• 1 mark — normal gradient = −1/2; equation y − 2 = −1/2 (x − 3), i.e. y = −x/2 + 7/2 (or 2y + x = 7).
Part (e) — 2 marks
• 1 mark — solves the simultaneous equations of the tangent and normal correctly, obtaining x = 3.
• 1 mark — confirms y = 2 at x = 3 in both the tangent and normal equations, stating the conclusion that they meet at (3, 2). Optional note: this is the same point as the point of tangency, which is geometrically obvious because the tangent and normal both pass through it.
Your response:
Stuck on (e)? The tangent and normal cross at the point of tangency — show this by substitution.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Derivative of f(x) = x² + 3x from first principles (3 marks)
Sample response. f(x + h) = (x + h)² + 3(x + h) = x² + 2xh + h² + 3x + 3h. f(x + h) − f(x) = (x² + 2xh + h² + 3x + 3h) − (x² + 3x) = 2xh + h² + 3h. Divide by h (h ≠ 0): (2xh + h² + 3h) / h = 2x + h + 3. f ′(x) = limh→0 (2x + h + 3) = 2x + 3.
Marking notes. 1 mark — correct f(x + h) with full expansion. 1 mark — correct simplification of numerator (cancels x² and 3x terms). 1 mark — factor h, cancel with "h ≠ 0" note, take limit, state derivative. Common errors: writing (x + h)² = x² + h² (no middle term); cancelling h before factoring out; missing the limit step.
1.2 — Equation of tangent to y = x³ at (1, 1) (3 marks)
Sample response. dy/dx = 3x². At x = 1: gradient = 3(1)² = 3. Point: (1, 1). Tangent: y − 1 = 3(x − 1) ⇒ y = 3x − 2.
Marking notes. 1 mark — correct gradient at x = 1 from quoted derivative. 1 mark — correct point of tangency (1, 1). 1 mark — final tangent equation in any acceptable form (e.g. y = 3x − 2 or 3x − y − 2 = 0). Common error: differentiating x³ as "x²" instead of 3x².
1.3 — Why d/dx (mx + c) = m (3 marks)
Sample response. The graph of f(x) = mx + c is a straight line of gradient m. The tangent to a straight line at any point on it is the line itself, because a straight line has the same direction at every point. Hence the gradient of the tangent — which is what the derivative measures — equals m at every x, so f ′(x) = m. (Algebraic check from first principles: ((m(x + h) + c) − (mx + c)) / h = mh/h = m for any h ≠ 0, and limh→0 m = m.)
Marking notes. 1 mark — clear statement that the tangent to a straight line is the line itself. 1 mark — connects "derivative = gradient of tangent" to "gradient of the line = m". 1 mark — concludes "for all x", i.e. constant function equal to m. Bare algebraic answer with no geometric reasoning scores 1.5/3 because the question asked for a geometric explanation.
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). Let f(x) = x² − 4x + 5. Then
f(x + h) = (x + h)² − 4(x + h) + 5 = x² + 2xh + h² − 4x − 4h + 5.
[1 mark — expansion.]
f(x + h) − f(x) = (x² + 2xh + h² − 4x − 4h + 5) − (x² − 4x + 5) = 2xh + h² − 4h.
[1 mark — numerator simplification, x², 4x and constant cancel.]
( f(x + h) − f(x) ) / h = h(2x + h − 4) / h = 2x + h − 4 (h ≠ 0).
f ′(x) = limh→0 (2x + h − 4) = 2x − 4. [1 mark — factor, cancel, limit.]
Part (b). Tangent parallel to x-axis ⇔ dy/dx = 0 ⇔ 2x − 4 = 0 ⇔ x = 2. y = (2)² − 4(2) + 5 = 4 − 8 + 5 = 1. The point is (2, 1). [1 mark.]
Part (c). At x = 3: gradient = 2(3) − 4 = 2; point (3, 9 − 12 + 5) = (3, 2). Tangent: y − 2 = 2(x − 3) ⇒ y = 2x − 4. [1 mark.]
Part (d). Normal gradient = −1/2 (negative reciprocal of 2). Through (3, 2): y − 2 = −1/2 (x − 3) ⇒ y = −x/2 + 7/2 (equivalently 2y + x = 7). [1 mark.]
Part (e). To find the intersection, solve y = 2x − 4 and 2y + x = 7 simultaneously. Substitute the first into the second: 2(2x − 4) + x = 7 ⇒ 4x − 8 + x = 7 ⇒ 5x = 15 ⇒ x = 3. [1 mark — correct simultaneous solution.] Then y = 2(3) − 4 = 2. Check in the normal: 2(2) + 3 = 7 ✓. Hence the tangent and normal meet at (3, 2), which is precisely the point of tangency — the only point common to both lines. [1 mark — verification and conclusion.]
Total: 8/8.
Band descriptors for marker.
Band 3: Attempts first principles in (a) but algebra incomplete; finds gradient at a point but cannot write the tangent equation correctly. ≈ 3-4 marks.
Band 4: Completes (a)-(c) but treats (d) and (e) procedurally without explaining the role of the negative reciprocal or the geometric meaning of the intersection. ≈ 5-6 marks.
Band 5: Completes (a)-(d); attempts (e) but does not explicitly tie the algebraic answer back to "the tangent and normal share only the point of tangency". ≈ 6-7 marks.
Band 6: Every step clean with proper limit notation, "h ≠ 0" annotated, algebra carefully checked, and (e) closed with the geometric observation that both lines must pass through the point of tangency. 8/8.