CIE AS/A Level Maths-3.4 Differentiation- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-3.4 Differentiation- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-3.4 Differentiation- Study Notes
Key Concepts:
- Derivatives of Common Functions
- Differentiation: Products and Quotients Rule
- Derivatives of Parametric and Implicit Functions
Derivatives of Common Functions
Derivatives of Common Functions
Basic Derivatives:
- \(\dfrac{d}{dx} e^x = e^x\)
- \(\dfrac{d}{dx} \ln x = \dfrac{1}{x}, \ x > 0\)
- \(\dfrac{d}{dx} \sin x = \cos x\)
- \(\dfrac{d}{dx} \cos x = -\sin x\)
- \(\dfrac{d}{dx} \tan x = \sec^2 x, \ x \neq \frac{\pi}{2} + n\pi\)
- \(\dfrac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}\)
Rules for Combination:
- Constant Multiple Rule: \(\dfrac{d}{dx}[k f(x)] = k f'(x)\)
- Sum/Difference Rule: \(\dfrac{d}{dx}[f(x) \pm g(x)] = f'(x) \pm g'(x)\)
- Composite/Chain Rule: \(\dfrac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)
Example
Find the derivative of \(y = 3 \sin x – 5 \ln x + 2 e^x\)
▶️ Answer / Explanation
Step 1: Apply derivative term by term:
\(\dfrac{dy}{dx} = 3 \dfrac{d}{dx}(\sin x) – 5 \dfrac{d}{dx}(\ln x) + 2 \dfrac{d}{dx}(e^x)\)
Step 2: Substitute derivatives:
\(\dfrac{dy}{dx} = 3 \cos x – \dfrac{5}{x} + 2 e^x\)
Final Answer: \(\boxed{\dfrac{dy}{dx} = 3 \cos x – \dfrac{5}{x} + 2 e^x}\)
Example
Find the derivative of \(y = \ln(5x^2 + 1)\)
▶️ Answer / Explanation
Step 1: Let \(u = 5x^2 + 1\), then \(y = \ln u\)
Step 2: Chain rule: \(\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} = \dfrac{1}{u} \cdot 10x\)
Step 3: Substitute back \(u = 5x^2 + 1\):
\(\dfrac{dy}{dx} = \dfrac{10x}{5x^2 + 1}\)
Final Answer: \(\boxed{\dfrac{dy}{dx} = \dfrac{10x}{5x^2 + 1}}\)
Example
Find \(\dfrac{d}{dx} [e^{2x} \tan^{-1}(x)]\)
▶️ Answer / Explanation
Step 1: Product rule: \(\dfrac{d}{dx}[f g] = f’ g + f g’\)
Here, \(f = e^{2x} \Rightarrow f’ = 2 e^{2x}\), \(g = \tan^{-1} x \Rightarrow g’ = \frac{1}{1+x^2}\)
Step 2: Apply product rule:
\(\dfrac{d}{dx}[e^{2x} \tan^{-1} x] = (2 e^{2x})(\tan^{-1} x) + (e^{2x})\left(\dfrac{1}{1+x^2}\right)\)
Final Answer: \(\boxed{\dfrac{d}{dx}[e^{2x} \tan^{-1} x] = 2 e^{2x} \tan^{-1} x + \dfrac{e^{2x}}{1+x^2}}\)
Example
Find \(\dfrac{d}{dx} [3 \sin x + 4 e^{x^2}]\)
▶️ Answer / Explanation
Step 1: Term by term derivative:
\(\dfrac{d}{dx}[3 \sin x] = 3 \cos x\)
\(\dfrac{d}{dx}[4 e^{x^2}] = 4 \cdot \dfrac{d}{dx} e^{x^2} = 4 \cdot e^{x^2} \cdot 2x = 8x e^{x^2}\)
Final Answer: \(\boxed{\dfrac{d}{dx}[3 \sin x + 4 e^{x^2}] = 3 \cos x + 8x e^{x^2}}\)
Differentiation: Products and Quotients
Differentiation: Products and Quotients
Rules:
- Product Rule: \(\dfrac{d}{dx}[f(x) g(x)] = f'(x) g(x) + f(x) g'(x)\)
- Quotient Rule: \(\dfrac{d}{dx}\left[\dfrac{f(x)}{g(x)}\right] = \dfrac{f'(x) g(x) – f(x) g'(x)}{[g(x)]^2}\)
- Chain rule may be required for composite functions.
Example
Differentiate \(y = \dfrac{x^3}{2x^2 + 4}\)
▶️ Answer / Explanation
Step 1: Identify \(f(x) = x^3, g(x) = 2x^2 + 4\)
Step 2: Compute derivatives:
\(f'(x) = 3x^2, \ g'(x) = 4x\)
Step 3: Apply quotient rule:
\(\dfrac{dy}{dx} = \dfrac{(3x^2)(2x^2+4) – (x^3)(4x)}{(2x^2 + 4)^2} = \dfrac{6x^4 + 12x^2 – 4x^4}{(2x^2 + 4)^2} = \dfrac{2x^4 + 12x^2}{(2x^2 + 4)^2}\)
Step 4: Factorize numerator:
\(\dfrac{dy}{dx} = \dfrac{2x^2(x^2 + 6)}{(2x^2 + 4)^2}\)
Final Answer: \(\boxed{\dfrac{dy}{dx} = \dfrac{2x^2(x^2 + 6)}{(2x^2 + 4)^2}}\)
Example
Differentiate \(y = x^2 \ln x\)
▶️ Answer / Explanation
Step 1: Identify \(f(x) = x^2, g(x) = \ln x\)
Step 2: Compute derivatives:
\(f'(x) = 2x, \ g'(x) = \frac{1}{x}\)
Step 3: Apply product rule:
\(\dfrac{dy}{dx} = (2x)(\ln x) + (x^2)\left(\frac{1}{x}\right) = 2x \ln x + x\)
Final Answer: \(\boxed{\dfrac{dy}{dx} = x(2 \ln x + 1)}\)
Example
Differentiate \(y = x e^{1 – x^2}\)
▶️ Answer / Explanation
Step 1: Identify \(f(x) = x, g(x) = e^{1 – x^2}\)
Step 2: Compute derivatives:
\(f'(x) = 1, \ g'(x) = e^{1 – x^2} \cdot (-2x) = -2x e^{1 – x^2}\)
Step 3: Apply product rule:
\(\dfrac{dy}{dx} = (1) e^{1 – x^2} + (x)(-2x e^{1 – x^2}) = e^{1 – x^2} – 2x^2 e^{1 – x^2} = e^{1 – x^2} (1 – 2x^2)\)
Final Answer: \(\boxed{\dfrac{dy}{dx} = e^{1 – x^2} (1 – 2x^2)}\)
Derivatives of Parametric and Implicit Functions
Derivatives of Parametric and Implicit Functions
Parametric Functions
If \(x = f(t)\) and \(y = g(t)\), then the derivative \(\dfrac{dy}{dx}\) is given by:
\(\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}, \quad \frac{dx}{dt} \neq 0\)
- Used for finding slopes of tangents and normals: slope of tangent = \(\dfrac{dy}{dx}\), slope of normal = \(-\dfrac{dx}{dy}\) or \(-1/(dy/dx)\)
Example
Given \(x = t – e^{2t}\), \(y = t + e^{2t}\). Find \(\dfrac{dy}{dx}\).
▶️ Answer / Explanation
Step 1: Differentiate both functions with respect to \(t\):
\(\dfrac{dx}{dt} = 1 – 2e^{2t}, \quad \dfrac{dy}{dt} = 1 + 2e^{2t}\)
Step 2: Apply formula \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\):
\(\dfrac{dy}{dx} = \dfrac{1 + 2 e^{2t}}{1 – 2 e^{2t}}\)
Step 3: Slope of tangent = \(\dfrac{1 + 2 e^{2t}}{1 – 2 e^{2t}}\)
Step 4: Slope of normal = \(- \dfrac{1 – 2 e^{2t}}{1 + 2 e^{2t}}\)
Final Answer: \(\boxed{\dfrac{dy}{dx} = \dfrac{1 + 2 e^{2t}}{1 – 2 e^{2t}}, \text{ normal slope } = -\dfrac{1 – 2 e^{2t}}{1 + 2 e^{2t}}}\)
Implicit Differentiation
- When \(y\) is defined implicitly by an equation \(F(x, y) = 0\), differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\) (\(dy/dx\) appears whenever differentiating a term involving \(y\))
- Solve for \(\dfrac{dy}{dx}\) to get the slope of the tangent
Example
Given \(x^2 + y^2 = xy + 7\), find \(\dfrac{dy}{dx}\).
▶️ Answer / Explanation
Step 1: Differentiate both sides with respect to \(x\):
\(\dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(y^2) = \dfrac{d}{dx}(xy) + \dfrac{d}{dx}(7)\)
Step 2: Apply derivatives (remember \(y = y(x)\)):
\(2x + 2y \dfrac{dy}{dx} = y + x \dfrac{dy}{dx} + 0\)
Step 3: Collect terms with \(\dfrac{dy}{dx}\) on LHS:
\(2y \dfrac{dy}{dx} – x \dfrac{dy}{dx} = y – 2x \Rightarrow (2y – x)\dfrac{dy}{dx} = y – 2x\)
Step 4: Solve for \(\dfrac{dy}{dx}\):
\(\dfrac{dy}{dx} = \dfrac{y – 2x}{2y – x}\)
Final Answer: \(\boxed{\dfrac{dy}{dx} = \dfrac{y – 2x}{2y – x}}\)
Use in Tangents and Normals:
- Slope of tangent = \(\dfrac{dy}{dx}\)
- Slope of normal = \(-1 / \dfrac{dy}{dx}\)
Example
Given a curve defined parametrically by \(x = t – e^{2t}, \ y = t + e^{2t}\). Find the equations of the tangent and normal at the point corresponding to \(t = 0\).
▶️ Answer / Explanation
Step 1: Differentiate \(x\) and \(y\) with respect to \(t\):
\(\dfrac{dx}{dt} = 1 – 2e^{2t}, \quad \dfrac{dy}{dt} = 1 + 2e^{2t}\)
Step 2: Find \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{1 + 2e^{2t}}{1 – 2e^{2t}}\)
Step 3: Evaluate slope at \(t = 0\):
\(\dfrac{dy}{dx}\Big|_{t=0} = \dfrac{1 + 2 \cdot 1}{1 – 2 \cdot 1} = \dfrac{3}{-1} = -3\)
Step 4: Find coordinates at \(t=0\):
\(x = 0 – e^0 = -1, \quad y = 0 + e^0 = 1\)
Step 5: Equation of tangent using point-slope form \(y – y_0 = m (x – x_0)\):
\(y – 1 = -3 (x + 1) \Rightarrow y – 1 = -3x – 3 \Rightarrow y = -3x – 2\)
Step 6: Equation of normal: slope = \(-1 / m = -1/(-3) = 1/3\)
\(y – 1 = \dfrac{1}{3}(x + 1) \Rightarrow y – 1 = \dfrac{1}{3}x + \dfrac{1}{3} \Rightarrow y = \dfrac{1}{3}x + \dfrac{4}{3}\)
Final Answer:
Tangent: \(\boxed{y = -3x – 2}\)
Normal: \(\boxed{y = \dfrac{1}{3}x + \dfrac{4}{3}}\)