CIE AS/A Level Maths-2.4 Differentiation- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-2.4 Differentiation- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-2.4 Differentiation- Study Notes
Key Concepts:
- Derivatives of Standard Functions
- Product Rule and Quotient Rule
- Differentiation of Parametric and Implicit Functions
Derivatives of Standard Functions
Derivatives of Standard Functions
Exponential and Logarithmic Functions
\(\dfrac{d}{dx}(e^x) = e^x\)
\(\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}, \; x>0\)
Trigonometric Functions
\(\dfrac{d}{dx}(\sin x) = \cos x\)
\(\dfrac{d}{dx}(\cos x) = -\sin x\)
\(\dfrac{d}{dx}(\tan x) = \sec^2 x, \; x \neq \dfrac{\pi}{2} + n\pi\)
Rules for Derivatives
(a) Constant Multiple Rule: \(\dfrac{d}{dx}[k f(x)] = k f'(x)\)
(b) Sum and Difference Rule: \(\dfrac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)\)
(c) Chain Rule (Composite Functions): If \(y = f(g(x))\), then \(\dfrac{dy}{dx} = f'(g(x)) \cdot g'(x)\)
Example:
Differentiate \(y = 5e^x – 3\ln x\).
▶️Answer/Explanation
Step 1: Differentiate each term separately.
\(\dfrac{d}{dx}(5e^x) = 5e^x\)
\(\dfrac{d}{dx}(-3\ln x) = -\dfrac{3}{x}\)
Step 2: Combine results.
\(\dfrac{dy}{dx} = 5e^x – \dfrac{3}{x}\)
Example:
Differentiate \(y = \sin(3x)\).
▶️Answer/Explanation
Step 1: Recognize composite function (\(\sin u\), where \(u=3x\)).
\(\dfrac{d}{dx}(\sin u) = \cos u \cdot \dfrac{du}{dx}\)
Step 2: Apply chain rule.
\(\dfrac{dy}{dx} = \cos(3x) \cdot 3\)
Final Answer: \(\dfrac{dy}{dx} = 3\cos(3x)\)
Example:
Differentiate \(y = x^2 \tan x\).
▶️Answer/Explanation
Step 1: Recognize product rule: \((uv)’ = u’v + uv’\).
Here \(u = x^2\), \(v = \tan x\).
Step 2: Differentiate each.
\(u’ = 2x\), \(v’ = \sec^2 x\).
Step 3: Apply product rule.
\(\dfrac{dy}{dx} = (2x)(\tan x) + (x^2)(\sec^2 x)\)
Product Rule and Quotient Rule
Product Rule
If \(y = u \cdot v\), where both \(u\) and \(v\) are functions of \(x\), then
\(\dfrac{dy}{dx} = u’v + uv’\)
Quotient Rule
If \(y = \dfrac{u}{v}\), where both \(u\) and \(v\) are functions of \(x\), then
\(\dfrac{dy}{dx} = \dfrac{u’v – uv’}{v^2}\)
Examples
Example
Differentiate \(y = x^2 e^x\).
▶️Answer/Explanation
Step 1: Identify \(u = x^2\), \(v = e^x\).
\(u’ = 2x\), \(v’ = e^x\).
Step 2: Apply product rule: \(y’ = u’v + uv’\).
\(y’ = (2x)(e^x) + (x^2)(e^x)\)
\(y’ = e^x(2x + x^2)\)
Example
Differentiate \(y = \dfrac{\sin x}{x}\).
▶️Answer/Explanation
Step 1: Identify \(u = \sin x\), \(v = x\).
\(u’ = \cos x\), \(v’ = 1\).
Step 2: Apply quotient rule: \(y’ = \dfrac{u’v – uv’}{v^2}\).
\(y’ = \dfrac{(\cos x)(x) – (\sin x)(1)}{x^2}\)
\(y’ = \dfrac{x \cos x – \sin x}{x^2}\)
Example
Differentiate \(y = \dfrac{x^2 \cos x}{e^x}\).
▶️Answer/Explanation
Step 1: Let \(u = x^2 \cos x\), \(v = e^x\).
First differentiate numerator using product rule.
\(u = x^2 \cos x\), so \(u’ = (2x \cos x – x^2 \sin x)\).
\(v = e^x\), so \(v’ = e^x\).
Step 2: Apply quotient rule.
\(y’ = \dfrac{u’v – uv’}{v^2}\)
= \(\dfrac{(2x\cos x – x^2 \sin x)(e^x) – (x^2 \cos x)(e^x)}{(e^x)^2}\)
Step 3: Simplify.
= \(\dfrac{e^x(2x\cos x – x^2 \sin x – x^2 \cos x)}{e^{2x}}\)
= \(\dfrac{2x \cos x – x^2 \sin x – x^2 \cos x}{e^x}\)
Differentiation of Parametric and Implicit Functions
Differentiation of Parametric Functions
When a curve is defined parametrically as
\(x = f(t), \; y = g(t)\),
then the derivative of \(y\) with respect to \(x\) is found by
\(\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\), provided \(\dfrac{dx}{dt} \neq 0\).
Example:
For \(x = t^2, \; y = t^3\), find \(\dfrac{dy}{dx}\).
▶️Answer/Explanation
Step 1: Differentiate both wrt \(t\).
\(\dfrac{dx}{dt} = 2t, \quad \dfrac{dy}{dt} = 3t^2\).
Step 2: Use the formula.
\(\dfrac{dy}{dx} = \dfrac{3t^2}{2t} = \dfrac{3t}{2}\).
Example:
The curve is given by \(x = \cos t, \; y = \sin t\). Find \(\dfrac{dy}{dx}\).
▶️Answer/Explanation
Step 1: Differentiate wrt \(t\).
\(\dfrac{dx}{dt} = -\sin t, \quad \dfrac{dy}{dt} = \cos t\).
Step 2: Apply the formula.
\(\dfrac{dy}{dx} = \dfrac{\cos t}{- \sin t} = -\cot t\).
Differentiation of Implicit Functions
Sometimes equations are given in terms of both \(x\) and \(y\) together (not solved explicitly for \(y\)). In such cases, differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\) (so use the chain rule).
Rule: If \(y\) appears, then \(\dfrac{d}{dx}[y] = \dfrac{dy}{dx}\).
Example:
Differentiate \(x^2 + y^2 = 25\) with respect to \(x\).
▶️Answer/Explanation
Step 1: Differentiate both sides.
\(\dfrac{d}{dx}[x^2 + y^2] = \dfrac{d}{dx}[25]\).
\(2x + 2y\dfrac{dy}{dx} = 0\).
Step 2: Solve for \(\dfrac{dy}{dx}\).
\(2y\dfrac{dy}{dx} = -2x\).
\(\dfrac{dy}{dx} = -\dfrac{x}{y}\).
Example:
Differentiate \(xy + y^2 = 10\) with respect to \(x\).
▶️Answer/Explanation
Step 1: Differentiate each term.
\(\dfrac{d}{dx}[xy] = x\dfrac{dy}{dx} + y\) (product rule, since \(y\) is a function of \(x\)).
\(\dfrac{d}{dx}[y^2] = 2y\dfrac{dy}{dx}\).
Step 2: Put together.
\(x\dfrac{dy}{dx} + y + 2y\dfrac{dy}{dx} = 0\).
Step 3: Factor.
\((x + 2y)\dfrac{dy}{dx} + y = 0\).
\(\dfrac{dy}{dx} = -\dfrac{y}{x+2y}\).