Source: ACE Exam Paper 1 ADV
Tags: differentiation, derivatives, product rule, quotient rule, chain rule, stationary points, curve sketching, tangent lines, optimisation, first derivative, second derivative, concavity
Differentiation is the backbone of this paper. You need fluency with the product rule, quotient rule, and chain rule, the ability to find and classify stationary points, and enough comfort with derivatives to handle applied optimisation problems. Most marks come from showing clean working, not just answers.
Derivative (f'(x) or dy/dx)
The rate of change of a function with respect to x. Geometrically, it gives the gradient of the tangent to the curve at any point.
Product rule
For y = uv: dy/dx = u(dv/dx) + v(du/dx). Used when two functions are multiplied together.
Quotient rule
For y = u/v: dy/dx = (v(du/dx) - u(dv/dx)) / v². Used when one function is divided by another.
Chain rule
For y = g(u) where u = f(x): dy/dx = (dy/du) × (du/dx). Used for composite functions (a function inside another function).
Stationary point
A point where f'(x) = 0. Could be a local maximum, local minimum, or horizontal point of inflection.
Second derivative (f''(x))
The derivative of the derivative. Tells you about concavity: f''(x) > 0 means concave up, f''(x) < 0 means concave down.
Concavity
The way a curve bends. Concave up looks like a cup (holds water). Concave down looks like a hill (sheds water).
Tangent line
A straight line that touches a curve at exactly one point and has the same gradient as the curve at that point.
When a graph is increasing (going uphill left to right), f'(x) > 0. When a graph is concave up (curving upward, getting steeper), f''(x) > 0. When a graph is concave down (curving and flattening), f''(x) < 0.
A curve that rises and flattens as it goes right has f'(x) > 0 (increasing) and f''(x) < 0 (concave down). This is the classic shape of a square root or logarithmic curve.
A curve that rises and gets steeper has f'(x) > 0 and f''(x) > 0.
f(x) = sin x + x² differentiates to f'(x) = cos x + 2x
f(x) = ln(x² + 1)³ uses the chain rule: f'(x) = 3 × (2x) / (x² + 1) = 6x / (x² + 1). The outer function is the cube, then the ln, then the inner (x² + 1). Alternatively, rewrite as 3 ln(x² + 1) first, then differentiate to get 3 × 2x/(x² + 1) = 6x/(x² + 1).
To differentiate 2eˣ cos x, let u = 2eˣ and v = cos x.
du/dx = 2eˣ
dv/dx = -sin x
dy/dx = 2eˣ cos x + 2eˣ(-sin x) = 2eˣ(cos x - sin x)
To differentiate (tan x)/x, let u = tan x and v = x.
du/dx = sec² x
dv/dx = 1
dy/dx = (x sec² x - tan x) / x²
For y = (x - a)(x² - 1), expand first: y = x³ - x - ax² + a.
dy/dx = 3x² - 2ax - 1
At x = -2: dy/dx = 3(4) - 2a(-2) - 1 = 12 + 4a - 1 = 4a + 11
The answer is (C) 4a + 11.
When you are given the graph of f'(x) and asked where f(x) is increasing, look for where the derivative graph is above the x-axis (f'(x) > 0).
If f'(x) is positive for -3 < x < 2 (the derivative curve sits above the x-axis in that interval), then f(x) is increasing on that interval.
For f(x) = x²(3 - x) = 3x² - x³:
f'(x) = 6x - 3x² = 3x(2 - x)
Setting f'(x) = 0: x = 0 or x = 2
f(0) = 0, so one stationary point is (0, 0)
f(2) = 4(3 - 2) = 4, so the other is (2, 4)
To determine nature, use the second derivative: f''(x) = 6 - 6x.
At x = 0: f''(0) = 6 > 0, so (0, 0) is a local minimum
At x = 2: f''(2) = 6 - 12 = -6 < 0, so (2, 4) is a local maximum
For f(x) = x²(3 - x) at the point P(1, 2):
f'(x) = 6x - 3x²
f'(1) = 6 - 3 = 3
Tangent: y - 2 = 3(x - 1), which gives y = 3x - 1
Given y = eᵏˣ sin x and (dy/dx) - 3y = eᵏˣ cos x:
By the product rule: dy/dx = keᵏˣ sin x + eᵏˣ cos x
Substituting into the equation: keᵏˣ sin x + eᵏˣ cos x - 3eᵏˣ sin x = eᵏˣ cos x
The eᵏˣ cos x terms cancel, leaving: keᵏˣ sin x - 3eᵏˣ sin x = 0
Factor: eᵏˣ sin x(k - 3) = 0
Since eᵏˣ sin x is not always zero, k = 3
For f(x) = 1/(1 + x²):
Rewrite as (1 + x²)⁻¹ or use the quotient rule
f'(x) = -2x / (1 + x²)²
To find where the tangent is parallel to the x-axis, set f'(x) = 0:
-2x / (1 + x²)² = 0
Numerator = 0, so x = 0
f(0) = 1/(1 + 0) = 1
The point is (0, 1)
A racetrack has two straights of length x and two semicircles with diameters 2y and y. Total perimeter = 4.8 km.
The perimeter equation: 2x + π(2y)/2 + π(y)/2 = 4.8, which simplifies to 2x + πy + πy/2 = 4.8, then 2x + 3πy/2 = 4.8.
Wait, the two semicircles have diameters 2y and y. Their circumferences are π(2y)/2 = πy and πy/2. So: 2x + πy + πy/2 = 4.8, giving 2x + 3πy/2 = 4.8. Rearranging: y = (9.6 - 4x) / (3π).
For speed S = 200 - (x³/27 + (π/6)y), substitute for y:
S = 200 - x³/27 - (π/6) × (9.6 - 4x)/(3π)
S = 200 - x³/27 - (9.6 - 4x)/18
Differentiate and set dS/dx = 0:
dS/dx = -3x²/27 + 4/18 = -x²/9 + 2/9
Setting to zero: x²/9 = 2/9, so x² = 2, giving x = √2
Verify it is a maximum: d²S/dx² = -2x/9, which is negative for x > 0. Confirmed maximum.
The length of the straight that maximises speed is √2 km (approximately 1.41 km).
Product rule: d/dx(uv) = u'v + uv'
Quotient rule: d/dx(u/v) = (vu' - uv') / v²
Chain rule: d/dx[f(g(x))] = f'(g(x)) × g'(x)
d/dx(sin x) = cos x
d/dx(cos x) = -sin x
d/dx(tan x) = sec² x
d/dx(eˣ) = eˣ
d/dx(ln x) = 1/x
d/dx(eᶠ⁽ˣ⁾) = f'(x) eᶠ⁽ˣ⁾
d/dx(ln f(x)) = f'(x)/f(x)
Tangent at point (x₁, y₁): y - y₁ = f'(x₁)(x - x₁)
⚠️ The product and quotient rules appear every year. Practise them until the setup is automatic.
⚠️ When finding the nature of stationary points, always show whether f''(x) is positive or negative at each point. A table of signs for f'(x) is an acceptable alternative.
⚠️ "Gradient at a point" means evaluate f'(x) at that x-value. Do not confuse gradient of the curve with gradient of a chord.
⚠️ In optimisation, always show that your answer is actually a maximum (or minimum) as required. A second derivative test or sign-change argument is expected.
⚠️ When differentiating ln(f(x))ⁿ, rewrite as n ln(f(x)) first. It is cleaner and avoids nested chain-rule errors.
⚠️ For "show that" questions, your working must arrive at the given expression. Do not start from the given expression and work backwards.
Q: Differentiate f(x) = sin x + x².
A: f'(x) = cos x + 2x
Q: Differentiate f(x) = ln(x² + 1)³.
A: Rewrite as 3 ln(x² + 1). Then f'(x) = 6x/(x² + 1).
Q: Find the gradient of y = (x - a)(x² - 1) at x = -2.
A: Expand to y = x³ - ax² - x + a. Differentiate: dy/dx = 3x² - 2ax - 1. At x = -2: dy/dx = 12 + 4a - 1 = 4a + 11.
Q: A graph is increasing and concave down. What are the signs of f'(x) and f''(x)?
A: f'(x) > 0 and f''(x) < 0.
Q: For f(x) = x²(3 - x), find and classify the stationary points.
A: f'(x) = 3x(2 - x) = 0 gives x = 0 and x = 2. f''(0) = 6 > 0 (local min at (0, 0)). f''(2) = -6 < 0 (local max at (2, 4)).
Q: Find the equation of the tangent to f(x) = x²(3 - x) at x = 1.
A: f'(1) = 3. Point is (1, 2). Tangent: y = 3x - 1.
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