Differentiation

1. Overview

Differentiation is a core topic in IGCSE Additional Mathematics, essential for understanding rates of change and optimization. It provides the tools to find gradients of curves, tangents, and normals, and to determine maximum and minimum values of functions. Mastery of differentiation is crucial for success in the 0606 exam, appearing in various contexts, from algebraic manipulation to practical problem-solving. This revision guide covers the key concepts, formulas, and techniques you need to excel.

Key Definitions

• Derivative: The function representing the gradient of the original curve, denoted as $\frac{dy}{dx}$ or $f'(x)$.
• Stationary Point: A point on a curve where the gradient is zero ($\frac{dy}{dx} = 0$), including local maxima and minima.
• Tangent: A straight line that touches a curve at a single point, sharing the same gradient as the curve at that point.
• Normal: A straight line perpendicular to the tangent at the point of contact.
• Chain Rule: A formula used to differentiate composite functions (a function within a function).
• Second Derivative: The derivative of the derivative ($f''(x)$), used to determine the nature of stationary points.

Core Content

A. Notation and the Derivative

The gradient of a curve $y = f(x)$ is the limit of the gradient of a chord as the change in $x$ ($\delta x$) approaches zero.

• Notation: $\frac{dy}{dx}$ (Leibniz notation), $f'(x)$ (Lagrange notation), and $\frac{d^2y}{dx^2}$ (second derivative).
• Delta Notation: $\frac{\delta y}{\delta x}$ represents a small change. As $\delta x \to 0$, $\frac{\delta y}{\delta x} \to \frac{dy}{dx}$.

B. Standard Derivatives

You must memorize the following (note: $a$ and $n$ are constants):

1. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
2. Exponential: $\frac{d}{dx}(e^{ax+b}) = ae^{ax+b}$
3. Logarithmic: $\frac{d}{dx}(\ln(ax+b)) = \frac{a}{ax+b}$
4. Trigonometric (Radians only):
   • $\frac{d}{dx}(\sin(ax+b)) = a\cos(ax+b)$
   • $\frac{d}{dx}(\cos(ax+b)) = -a\sin(ax+b)$
   • $\frac{d}{dx}(\tan(ax+b)) = a\sec^2(ax+b)$

C. Rules of Differentiation

• Chain Rule: Used for $y = f(g(x))$. $$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$
• Product Rule: Used for $y = uv$. $$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$
• Quotient Rule: Used for $y = \frac{u}{v}$. $$\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$

Worked Example 1 — Product Rule with Radian Trig

Differentiate $y = x^2 \sin(3x)$ with respect to $x$.

1. Identify $u$ and $v$: $u = x^2$, $v = \sin(3x)$ Reason: Prepare to apply the product rule.
2. Differentiate $u$ and $v$: $\frac{du}{dx} = 2x$, $\frac{dv}{dx} = 3\cos(3x)$ Reason: Apply power rule and trig derivative.
3. Apply the product rule: $\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx} = x^2(3\cos(3x)) + \sin(3x)(2x)$ Reason: Substitute into the product rule formula.
4. Simplify: $\frac{dy}{dx} = 3x^2\cos(3x) + 2x\sin(3x)$ Reason: Factor out common terms.
5. Final Answer: $\frac{dy}{dx} = x(3x\cos(3x) + 2\sin(3x))$ Reason: Fully simplified form.

Final Answer: $\boxed{\frac{dy}{dx} = x(3x\cos(3x) + 2\sin(3x))}$

Worked Example 2 — Quotient Rule with Logarithms

Differentiate $y = \frac{\ln x}{x^3}$ with respect to $x$.

1. Identify $u$ and $v$: $u = \ln x$, $v = x^3$ Reason: Prepare to apply the quotient rule.
2. Differentiate $u$ and $v$: $\frac{du}{dx} = \frac{1}{x}$, $\frac{dv}{dx} = 3x^2$ Reason: Apply standard derivatives.
3. Apply the quotient rule: $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} = \frac{x^3(\frac{1}{x}) - (\ln x)(3x^2)}{(x^3)^2}$ Reason: Substitute into the quotient rule formula.
4. Simplify: $\frac{dy}{dx} = \frac{x^2 - 3x^2\ln x}{x^6}$ Reason: Simplify the numerator and denominator.
5. Further simplification: $\frac{dy}{dx} = \frac{x^2(1 - 3\ln x)}{x^6} = \frac{1 - 3\ln x}{x^4}$ Reason: Cancel out common factors.

Final Answer: $\boxed{\frac{dy}{dx} = \frac{1 - 3\ln x}{x^4}}$

D. Tangents and Normals

• Gradient of Tangent ($m_t$): Find $\frac{dy}{dx}$ at $x = a$.
• Gradient of Normal ($m_n$): $m_n = -\frac{1}{m_t}$.
• Equation: Use $y - y_1 = m(x - x_1)$.

Worked Example 3 — Tangent and Normal with Exponential

Find the equation of the tangent and normal to the curve $y = e^{3x}$ at the point where $x = 0$.

1. Find the $y$-coordinate: $y = e^{3(0)} = e^0 = 1$ Reason: Substitute $x=0$ into the original equation. The point is $(0, 1)$.
2. Differentiate: $\frac{dy}{dx} = 3e^{3x}$ Reason: Apply the exponential derivative rule and chain rule.
3. Find the gradient of the tangent ($m_t$) at $x = 0$: $m_t = 3e^{3(0)} = 3e^0 = 3$ Reason: Substitute $x=0$ into the derivative.
4. Find the gradient of the normal ($m_n$): $m_n = -\frac{1}{m_t} = -\frac{1}{3}$ Reason: The normal is perpendicular to the tangent.
5. Equation of the tangent: $y - 1 = 3(x - 0)$ Reason: Use the point-slope form of a line. $y = 3x + 1$
6. Equation of the normal: $y - 1 = -\frac{1}{3}(x - 0)$ Reason: Use the point-slope form of a line. $y = -\frac{1}{3}x + 1$

Final Answer (Tangent): $\boxed{y = 3x + 1}$ Final Answer (Normal): $\boxed{y = -\frac{1}{3}x + 1}$

E. Stationary Points and the Second Derivative Test

To find stationary points, set $\frac{dy}{dx} = 0$ and solve for $x$. To determine the nature:

• If $\frac{d^2y}{dx^2} > 0$, the point is a Minimum.
• If $\frac{d^2y}{dx^2} < 0$, the point is a Maximum.

Worked Example 4 — Stationary Points and Nature

Find the stationary points of the curve $y = x^3 - 6x^2 + 9x$ and determine their nature.

1. Find the first derivative: $\frac{dy}{dx} = 3x^2 - 12x + 9$ Reason: Apply the power rule.
2. Set the first derivative to zero and solve for $x$: $3x^2 - 12x + 9 = 0$ Reason: Stationary points occur where the gradient is zero. $x^2 - 4x + 3 = 0$ Reason: Divide by 3. $(x - 1)(x - 3) = 0$ Reason: Factorize the quadratic. $x = 1$ or $x = 3$
3. Find the corresponding $y$-values: When $x = 1$, $y = (1)^3 - 6(1)^2 + 9(1) = 1 - 6 + 9 = 4$ When $x = 3$, $y = (3)^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0$ Reason: Substitute $x$ values into the original equation. The stationary points are $(1, 4)$ and $(3, 0)$.
4. Find the second derivative: $\frac{d^2y}{dx^2} = 6x - 12$ Reason: Differentiate the first derivative.
5. Determine the nature of the stationary points: At $x = 1$, $\frac{d^2y}{dx^2} = 6(1) - 12 = -6 < 0$. Therefore, $(1, 4)$ is a maximum point. At $x = 3$, $\frac{d^2y}{dx^2} = 6(3) - 12 = 6 > 0$. Therefore, $(3, 0)$ is a minimum point.

Final Answer (Stationary Points): $\boxed{(1, 4) \text{ (Maximum)}, (3, 0) \text{ (Minimum)}}$

F. Connected Rates of Change and Small Increments

• Connected Rates: $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$.
• Small Increments: $\delta y \approx \left(\frac{dy}{dx}\right) \times \delta x$.

Worked Example 5 — Connected Rates of Change

The radius $r$ of a circle is increasing at a rate of $0.2$ cm/s. Find the rate at which the area $A$ of the circle is increasing when $r = 5$ cm.

1. Identify the given rates: $\frac{dr}{dt} = 0.2$ cm/s Reason: This is the rate of change of the radius with respect to time.
2. Write the formula for the area of a circle: $A = \pi r^2$ Reason: This relates the area to the radius.
3. Differentiate $A$ with respect to $r$: $\frac{dA}{dr} = 2\pi r$ Reason: Apply the power rule.
4. Apply the chain rule (connected rates): $\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$ Reason: This connects the rate of change of area to the rate of change of radius.
5. Substitute the given values: $\frac{dA}{dt} = (2\pi (5)) \times (0.2) = 2\pi$ Reason: Substitute $r=5$ and $\frac{dr}{dt} = 0.2$ into the equation.

Final Answer: $\boxed{\frac{dA}{dt} = 2\pi \text{ cm}^2\text{/s}}$

Extended Content (Extended Only)

Additional Mathematics is a single-tier syllabus — all content above applies to all students.

Key Equations

Name	Formula	Notes
Product Rule	$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$	Given on formula sheet
Quotient Rule	$\frac{d}{dx}(\frac{u}{v}) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$	Given on formula sheet
Chain Rule	$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$	Not on sheet
Small Changes	$\delta y \approx f'(x) \cdot \delta x$	Use for approximations
Trig Derivatives	$\sin x \to \cos x, \cos x \to -\sin x$	Must be in Radians

Common Mistakes to Avoid

• ❌ Wrong: Calculating derivatives of trig functions using degrees. ✓ Right: Always ensure your calculator is in Radians mode and use $\pi$ for exact values.
• ❌ Wrong: Differentiating $\ln(f(x))$ as $\frac{1}{f(x)}$. ✓ Right: Remember the chain rule: $\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$.
• ❌ Wrong: In the quotient rule, reversing the numerator: $u\frac{dv}{dx} - v\frac{du}{dx}$. ✓ Right: It is always "bottom times derivative of top" first: $v\frac{du}{dx} - u\frac{dv}{dx}$.
• ❌ Wrong: Forgetting to differentiate the $2x$ in $e^{2x}$ or $\sin(2x)$. ✓ Right: Always multiply by the derivative of the "inner" function (Chain Rule).
• ❌ Wrong: Giving a decimal approximation when an exact value (surd or ln) is required. ✓ Right: Leave your answer in terms of $\sqrt{ }$, $\ln$, $\pi$, or fractions unless the question specifically asks for a decimal approximation.
• ❌ Wrong: Making sign errors when expanding brackets or applying the quotient rule. ✓ Right: Double-check your algebraic manipulation, especially when dealing with negative signs. Use brackets carefully.
• ❌ Wrong: Forgetting the domain restrictions of logarithmic functions. ✓ Right: Remember that $\ln(x)$ is only defined for $x > 0$. Check that your solutions are valid.
• ❌ Wrong: Missing solutions when solving trigonometric equations after differentiation. ✓ Right: Consider the periodicity of trigonometric functions and find all solutions within the given range.

Exam Tips

• Show Every Step: In "Show that" questions, never skip steps. Write the formula, show the substitution, and then the simplification.
• Exact Values: Unless the question asks for 3 significant figures, leave answers in exact form: e.g., $2\sqrt{3}$, $\frac{1}{2}\ln 5$, or $4\pi$.
• Stationary Point Justification: When asked to "determine the nature," you must show the value of the second derivative and explicitly state "$\because \frac{d^2y}{dx^2} < 0$, the point is a maximum."
• Command Words:
  • Find the rate of change: This usually means find $\frac{dy}{dt}$.
  • Find the gradient: Find $\frac{dy}{dx}$ and substitute the $x$ value.
  • Stationary value: This is the $y$-coordinate of the stationary point, not just the $x$ value.
• Calculator: On Paper 2, use the $\frac{d}{dx}$ button to verify your numerical gradient at a specific point, but you must show algebraic working for full marks. On Paper 1 (non-calculator), focus on mastering algebraic differentiation techniques.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0606 papers.

Exam-Style Question 1 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

The curve $y = \frac{3x}{\sqrt{x+2}}$ is defined for $x > -2$.

(a) Show that $\frac{dy}{dx} = \frac{3x+12}{2(x+2)^{\frac{3}{2}}}$. [4]

Worked Solution:

(a)

1. Apply the quotient rule: Let $u = 3x$ and $v = (x+2)^{\frac{1}{2}}$. Then $\frac{du}{dx} = 3$ and $\frac{dv}{dx} = \frac{1}{2}(x+2)^{-\frac{1}{2}}$. $\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} = \frac{3(x+2)^{\frac{1}{2}} - 3x(\frac{1}{2}(x+2)^{-\frac{1}{2}})}{(x+2)}$ [M1: Correct application of the quotient rule formula]

2. Simplify the expression: Multiply numerator and denominator by $(x+2)^{\frac{1}{2}}$. $\frac{dy}{dx} = \frac{3(x+2) - \frac{3}{2}x}{(x+2)^{\frac{3}{2}}} = \frac{3x+6-\frac{3}{2}x}{(x+2)^{\frac{3}{2}}}$ [M1: Multiplying to remove the negative power]

3. Further simplification: $\frac{dy}{dx} = \frac{\frac{3}{2}x+6}{(x+2)^{\frac{3}{2}}} = \frac{3x+12}{2(x+2)^{\frac{3}{2}}}$ [A1: Simplification of the numerator]

4. State the final answer: $\frac{dy}{dx} = \frac{3x+12}{2(x+2)^{\frac{3}{2}}}$ [A1: Correct final answer]

How to earn full marks: Show the quotient rule formula, then carefully substitute each term and simplify step-by-step to arrive at the given answer.

(b) Find the $x$-coordinate of the stationary point of the curve. [3]

Worked Solution:

(b)

1. For a stationary point, $\frac{dy}{dx} = 0$. Therefore, $\frac{3x+12}{2(x+2)^{\frac{3}{2}}} = 0$. [M1: Setting the derivative to zero]

2. Solve for $x$: This implies $3x+12 = 0$. $3x = -12$ $x = -4$ [A1: Finding the x-coordinate]

3. State the final answer: $x = -4$ [A1: Correct final answer]

How to earn full marks: Set the derivative equal to zero, solve the resulting equation, and clearly state the x-coordinate of the stationary point.

Common Pitfall: Remember to apply the quotient rule carefully, paying close attention to the order of terms in the numerator. Also, don't forget to fully simplify your expression after applying the rule.

Exam-Style Question 2 — Paper 1 (No Calculator Allowed) [7 marks]

Question:

The function $f(x)$ is defined by $f(x) = (2x-1)^2(x+1)$.

(a) Find $f'(x)$. [4]

Worked Solution:

(a)

1. Apply the product rule: Let $u = (2x-1)^2$ and $v = (x+1)$. Then $\frac{du}{dx} = 2(2x-1)(2) = 4(2x-1)$ and $\frac{dv}{dx} = 1$. $\frac{d}{dx}((2x-1)^2(x+1)) = (2x-1)^2(1) + (x+1)(4(2x-1))$ [M1: Correct application of the product rule and chain rule]

2. Simplify the expression: $f'(x) = (2x-1)^2 + 4(x+1)(2x-1) = (2x-1)[(2x-1) + 4(x+1)]$ [M1: Factoring out (2x-1)]

3. Further simplification: $f'(x) = (2x-1)(2x-1+4x+4) = (2x-1)(6x+3)$ [A1: Correct algebraic manipulation]

4. State the final answer: $f'(x) = (2x-1)(6x+3)$ [A1: Correct final answer]

How to earn full marks: Apply both the product and chain rules correctly, then simplify by factoring to get the derivative in a manageable form.

(b) Find the values of $x$ for which $f'(x) = 0$. [3]

Worked Solution:

(b)

1. Set $f'(x) = 0$: $(2x-1)(6x+3) = 0$. [M1: Setting the derivative to zero]

2. Solve for $x$: $2x-1 = 0$ or $6x+3 = 0$. $x = \frac{1}{2}, x = -\frac{1}{2}$ [A1: Finding both x-values]

3. State the final answer: $x = \frac{1}{2}, -\frac{1}{2}$ [A1: Correct final answer]

How to earn full marks: Set the derivative to zero, factorize if necessary, and solve for all possible values of x.

Common Pitfall: When using the product rule, make sure you correctly differentiate each part of the product, including applying the chain rule where necessary. Factoring out common terms can simplify the expression and make it easier to solve.

Exam-Style Question 3 — Paper 2 (Calculator Allowed) [8 marks]

Question:

A curve has the equation $y = x^3 - 4x^2 + 2x + 5$.

(a) Find the coordinates of the stationary points of the curve. [5]

Worked Solution:

(a)

1. Find the first derivative: $\frac{dy}{dx} = 3x^2 - 8x + 2$ [B1: Correct differentiation]

2. Set the derivative to zero: $3x^2 - 8x + 2 = 0$. [M1: Setting the derivative to zero]

3. Solve the quadratic equation: Using the quadratic formula: $x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(2)}}{2(3)} = \frac{8 \pm \sqrt{64 - 24}}{6} = \frac{8 \pm \sqrt{40}}{6} = \frac{8 \pm 2\sqrt{10}}{6} = \frac{4 \pm \sqrt{10}}{3}$ $x = \frac{4 + \sqrt{10}}{3} \approx 2.387, x = \frac{4 - \sqrt{10}}{3} \approx 0.279$ [M1: Solving the quadratic equation]

4. Find the corresponding $y$-coordinates: When $x = 2.387$, $y = (2.387)^3 - 4(2.387)^2 + 2(2.387) + 5 \approx -0.069$. When $x = 0.279$, $y = (0.279)^3 - 4(0.279)^2 + 2(0.279) + 5 \approx 5.280$. [A1: Finding the y-coordinate for each x value]

5. State coordinates: $(2.387, -0.069)$, $(0.279, 5.280)$ [A1: Correct final answer]

How to earn full marks: Differentiate, set to zero, use the quadratic formula correctly, and find both x and y coordinates for each stationary point.

(b) Determine the nature of each stationary point. [3]

Worked Solution:

(b)

1. Find the second derivative: $\frac{d^2y}{dx^2} = 6x - 8$ [M1: Correct second derivative]

2. Evaluate the second derivative at each stationary point: When $x = 2.387$, $\frac{d^2y}{dx^2} = 6(2.387) - 8 \approx 6.322 > 0$. Therefore, it is a minimum point. When $x = 0.279$, $\frac{d^2y}{dx^2} = 6(0.279) - 8 \approx -6.326 < 0$. Therefore, it is a maximum point. [A1: Correctly finding the value and sign of the second derivative]

3. State the nature of the stationary points: $(2.387, -0.069)$ is a minimum point. $(0.279, 5.280)$ is a maximum point. [A1: Correctly stating the nature of each point]

How to earn full marks: Find the second derivative, substitute each x-coordinate, and correctly interpret the sign to determine if it's a max or min.

Common Pitfall: Remember to use the quadratic formula correctly when you can't easily factorize the quadratic equation. Also, be careful with signs when evaluating the second derivative to determine the nature of the stationary points.

Exam-Style Question 4 — Paper 2 (Calculator Allowed) [8 marks]

Question:

A particle moves along the curve $y = \ln(2x^2 + 1)$ for $x > 0$. The $x$-coordinate of the particle is changing at a rate of 0.3 units per second.

(a) Find the rate of change of the $y$-coordinate with respect to time when $x = 1$. [5]

Worked Solution:

(a)

1. Differentiate $y$ with respect to $x$: $\frac{dy}{dx} = \frac{4x}{2x^2+1}$ [M1: Correct application of the chain rule] [A1: Correct derivative]

2. State the given rate: $\frac{dx}{dt} = 0.3$. [B1: Correct rate]

3. Apply the chain rule: $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}$. $\frac{dy}{dt} = \frac{4x}{2x^2+1} \times 0.3$ [M1: Correctly using chain rule]

4. Substitute $x = 1$: $\frac{dy}{dt} = \frac{4(1)}{2(1)^2+1} \times 0.3 = \frac{4}{3} \times 0.3 = 0.4$ [M1: Substitute and calculate value]

5. State the final answer: The rate of change of the $y$-coordinate is 0.4 units per second. $\boxed{0.4}$ [A1: Correct final answer]

How to earn full marks: Correctly apply the chain rule, identify the given rate, and substitute the value of x to find the rate of change of y.

(b) Find the equation of the tangent to the curve at the point where $x = 1$. Give your answer in the form $y = mx + c$, where $m$ and $c$ are correct to 3 significant figures. [3]

Worked Solution:

(b)

1. Find the $y$-coordinate when $x = 1$: $y = \ln(2(1)^2 + 1) = \ln(3) \approx 1.099$ [M1: Correct y-coordinate]

2. Find the gradient of the tangent when $x = 1$: $\frac{dy}{dx} = \frac{4(1)}{2(1)^2+1} = \frac{4}{3} \approx 1.333$ [A1: Correct gradient]

3. Use the point-slope form of a line: $y - y_1 = m(x - x_1)$. $y - \ln(3) = \frac{4}{3}(x - 1)$. $y = \frac{4}{3}x - \frac{4}{3} + \ln(3)$. $y = 1.333x - 1.333 + 1.099$ $y = 1.333x - 0.234$ State the equation: $y = 1.33x - 0.234$ [A1: Correct equation of tangent]

How to earn full marks: Find the y-coordinate and the gradient at x=1, then use the point-slope form to find the tangent equation, rounding only at the end.

Common Pitfall: When dealing with related rates, remember to correctly apply the chain rule. Also, ensure you substitute the given value of x at the appropriate step in the calculation. When finding the tangent equation, remember to round only the final answer to the specified significant figures.