How to Solve: Differentiation (First Principles, Product/Quotient/Chain Rules, Parametric, Implicit)

How to Solve: Differentiation (First Principles, Product/Quotient/Chain Rules, Parametric, Implicit)

GCSE / A-Level Maths

Introduction

"Differentiation is the engine behind every optimisation problem—whether you’re designing a rocket, maximising profit, or just acing your A-Level exam. Master these rules, and you’ll unlock 15-20% of your final grade in calculus questions. Let’s break it down so you never freeze on exam day."

What You Need To Know First

1. Basic algebra – Rearranging equations, expanding brackets, simplifying fractions.
2. Limits – Understanding what h → 0 means (for first principles).
3. Function notation – Knowing f(x), f'(x), and how to substitute values.

Key Vocabulary

Formulas To Know

1. First Principles (Definition of the Derivative)

Formula: f'(x) = lim(h→0) [f(x+h) – f(x)] / h - f(x) = original function - h = small change in x - MEMORISE THIS (but rarely used directly in exams—just for understanding).

2. Power Rule

Formula: d/dx [xⁿ] = n xⁿ⁻¹ - n = any real number (integer, fraction, negative). - MEMORISE THIS (most common rule).

3. Product Rule

Formula: d/dx [u·v] = u’v + uv’ - u and v = two functions of x. - u’ = derivative of u, v’ = derivative of v. - MEMORISE THIS.

4. Quotient Rule

Formula: d/dx [u/v] = (u’v – uv’) / v² - u = numerator, v = denominator. - MEMORISE THIS (or derive from product rule if stuck).

5. Chain Rule

Formula: d/dx [f(g(x))] = f’(g(x)) · g’(x) - f = outer function, g = inner function. - MEMORISE THIS.

6. Parametric Differentiation

Formula: dy/dx = (dy/dt) / (dx/dt) - x and y are both functions of t. - MEMORISE THIS.

7. Implicit Differentiation

Method: Differentiate both sides w.r.t. x, treating y as a function of x (so d/dx [y] = dy/dx). - MEMORISE THIS (no single formula—just a process).

Step-by-Step Method

1. First Principles (Definition)

When to use: When asked to "prove from first principles" or "use the definition of the derivative." Steps: 1. Write the definition: f'(x) = lim(h→0) [f(x+h) – f(x)] / h. 2. Substitute f(x) and f(x+h) into the formula. 3. Expand and simplify the numerator. 4. Factor out h from the numerator. 5. Cancel h in the numerator and denominator. 6. Take the limit as h → 0 (substitute h = 0). 7. Simplify to get f'(x).

Worked Example: Find the derivative of f(x) = x² from first principles. 1. f'(x) = lim(h→0) [(x+h)² – x²] / h 2. Expand: [(x² + 2xh + h²) – x²] / h = (2xh + h²) / h 3. Factor: h(2x + h) / h = 2x + h 4. Limit: lim(h→0) (2x + h) = 2x Answer: f'(x) = 2x

2. Product Rule

When to use: When differentiating u(x) · v(x) (two functions multiplied). Steps: 1. Identify u and v. 2. Find u’ and v’ (derivatives of u and v). 3. Apply the formula: u’v + uv’. 4. Simplify the result.

Worked Example: Differentiate f(x) = x² sin(x). 1. u = x² → u’ = 2x 2. v = sin(x) → v’ = cos(x) 3. f'(x) = u’v + uv’ = 2x sin(x) + x² cos(x) Answer: f'(x) = 2x sin(x) + x² cos(x)

3. Quotient Rule

When to use: When differentiating u(x) / v(x) (a fraction). Steps: 1. Identify u (numerator) and v (denominator). 2. Find u’ and v’ (derivatives of u and v). 3. Apply the formula: (u’v – uv’) / v². 4. Simplify the result (factor if possible).

Worked Example: Differentiate f(x) = (3x + 1) / (x² – 2). 1. u = 3x + 1 → u’ = 3 2. v = x² – 2 → v’ = 2x 3. f'(x) = [3(x² – 2) – (3x + 1)(2x)] / (x² – 2)² 4. Expand numerator: 3x² – 6 – 6x² – 2x = -3x² – 2x – 6 5. Factor numerator: - (3x² + 2x + 6) Answer: f'(x) = - (3x² + 2x + 6) / (x² – 2)²

4. Chain Rule

When to use: When differentiating a composite function f(g(x)) (a function inside another). Steps: 1. Identify the inner function g(x) and outer function f(u) (where u = g(x)). 2. Differentiate the outer function w.r.t. u: f’(u). 3. Differentiate the inner function w.r.t. x: g’(x). 4. Multiply: f’(g(x)) · g’(x). 5. Simplify.

Worked Example: Differentiate f(x) = sin(3x² + 1). 1. Inner: g(x) = 3x² + 1 → g’(x) = 6x 2. Outer: f(u) = sin(u) → f’(u) = cos(u) 3. Apply chain rule: f'(x) = cos(3x² + 1) · 6x Answer: f'(x) = 6x cos(3x² + 1)

5. Parametric Differentiation

When to use: When x and y are both given in terms of a parameter t. Steps: 1. Find dx/dt and dy/dt. 2. Divide: dy/dx = (dy/dt) / (dx/dt). 3. Simplify if possible.

Worked Example: Given x = t² + 1, y = 2t³, find dy/dx. 1. dx/dt = 2t 2. dy/dt = 6t² 3. dy/dx = (6t²) / (2t) = 3t Answer: dy/dx = 3t

6. Implicit Differentiation

When to use: When y is not isolated (e.g., x² + y² = 1). Steps: 1. Differentiate both sides w.r.t. x. 2. Treat y as a function of x (so d/dx [y] = dy/dx). 3. Use chain rule for terms with y (e.g., d/dx [y²] = 2y dy/dx). 4. Collect dy/dx terms on one side. 5. Factor out dy/dx and solve.

Worked Example: Find dy/dx for x² + y² = 25. 1. Differentiate: 2x + 2y dy/dx = 0 2. Solve for dy/dx: 2y dy/dx = -2x 3. dy/dx = -2x / 2y = -x/y Answer: dy/dx = -x/y

Worked Examples

Example 1 – Basic (Chain Rule)

Differentiate f(x) = (5x + 2)⁴. 1. Inner: g(x) = 5x + 2 → g’(x) = 5 2. Outer: f(u) = u⁴ → f’(u) = 4u³ 3. Chain rule: f'(x) = 4(5x + 2)³ · 5 = 20(5x + 2)³ What we did and why: Recognised a composite function, applied chain rule, and simplified.

Example 2 – Medium (Product + Chain Rule)

Differentiate f(x) = x² e^(3x). 1. u = x² → u’ = 2x 2. v = e^(3x) → v’ = 3e^(3x) (chain rule) 3. Product rule: f'(x) = u’v + uv’ = 2x e^(3x) + x² · 3e^(3x) = e^(3x)(2x + 3x²) What we did and why: Combined product and chain rules, factored out common terms.

Example 3 – Exam-Style (Implicit + Quotient Rule)

Given x² y + y² = 4, find dy/dx at (1, 2). 1. Differentiate implicitly: 2x y + x² dy/dx + 2y dy/dx = 0 2. Collect dy/dx: dy/dx (x² + 2y) = -2x y 3. Solve: dy/dx = -2x y / (x² + 2y) 4. Substitute (1, 2): dy/dx = -2(1)(2) / (1 + 4) = -4/5 What we did and why: Used implicit differentiation, isolated dy/dx, and substituted the point.

Common Mistakes

Exam Traps

1-Minute Recap

"Right, listen up—this is your 60-second survival guide for differentiation. First principles? Only if the question says so—otherwise, use the rules. Product rule: u’v + uv’. Quotient rule: (u’v – uv’) / v². Chain rule: derivative of the outside times derivative of the inside. Parametric? Divide dy/dt by dx/dt. Implicit? Differentiate everything, treat y like a function, and solve for dy/dx. Common mistakes? Forgetting dy/dx, mixing up signs, and not simplifying. Exam traps? Watch for hidden chain rule and implicit products. Now go practice—you’ve got this!