Pure 1: Differentiation

Pure 1: Differentiation — CIE A Level Mathematics 9709

1. Differentiation from First Principles

The derivative of f(x) is defined as:

$$f'(x) = lim_{h o 0} rac{f(x+h) - f(x)}{h}$$

Example: Differentiate f(x) = x² from first principles.

f(x+h) = (x+h)² = x² + 2xh + h²

f'(x) = lim[(x² + 2xh + h² − x²)/h] = lim[2x + h] = 2x ✓

2. Standard Results

| f(x) | f'(x) |

|---|---|

| xⁿ | nxⁿ⁻¹ |

| sin x | cos x |

| cos x | −sin x |

| tan x | sec²x |

| eˣ | eˣ |

| ln x | 1/x |

| aˣ | aˣ ln a |

Power rule: d/dx(xⁿ) = nxⁿ⁻¹ (works for fractional and negative powers too)

3. Rules of Differentiation

Sum/difference rule: d/dx[f(x) ± g(x)] = f'(x) ± g'(x)

Chain rule: For y = f(g(x)): dy/dx = f'(g(x)) × g'(x)

Or: if y = uⁿ, dy/dx = nuⁿ⁻¹ × du/dx

Example: y = (3x² + 1)⁵

dy/dx = 5(3x²+1)⁴ × 6x = 30x(3x² + 1)⁴

Product rule: d/dx[uv] = u(dv/dx) + v(du/dx)

Example: y = x²sin x

dy/dx = x²(cos x) + sin x(2x) = x²cos x + 2x sin x

Quotient rule: d/dx[u/v] = [v(du/dx) − u(dv/dx)] / v²

Example: y = sin x / (x² + 1)

dy/dx = [(x²+1)cos x − sin x(2x)] / (x²+1)²

4. Implicit Differentiation

Differentiate each term with respect to x, using chain rule for y terms (multiply by dy/dx).

Example: Differentiate x² + y² = 25 implicitly.

2x + 2y(dy/dx) = 0 → dy/dx = −x/y

At (3, 4): dy/dx = −3/4 (gradient of tangent at this point).

5. Stationary Points

At a stationary point: dy/dx = 0.

Second derivative test:

• d²y/dx² > 0: minimum
• d²y/dx² < 0: maximum
• d²y/dx² = 0: inconclusive — check by looking at sign of dy/dx on either side

Example: y = x³ − 3x² − 9x + 5. Find and classify stationary points.

dy/dx = 3x² − 6x − 9 = 3(x² − 2x − 3) = 3(x − 3)(x + 1) = 0

x = 3 or x = −1.

d²y/dx² = 6x − 6.

• At x = 3: d²y/dx² = 12 > 0 → minimum at (3, −22)
• At x = −1: d²y/dx² = −12 < 0 → maximum at (−1, 10)

6. Applications: Optimisation

Method: Write the quantity to be optimised (area, volume, cost) as a function of one variable. Differentiate, set = 0, verify it's a max/min.

Example: An open box with square base of side x cm and height h cm has volume 32 cm³. The base costs PKR 3/cm² and sides PKR 1/cm². Minimise cost.

• Constraint: x²h = 32 → h = 32/x²
• Cost: C = 3x² + 4xh = 3x² + 4x(32/x²) = 3x² + 128/x
• dC/dx = 6x − 128/x² = 0 → x³ = 128/6 → x ≈ 2.75 cm

7. Connected Rates of Change

Using chain rule: dy/dt = (dy/dx) × (dx/dt)

Example: A spherical balloon is inflated at 50 cm³/s. When radius r = 5 cm, find dr/dt.

V = (4/3)πr³ → dV/dr = 4πr²

dV/dt = (dV/dr)(dr/dt) → 50 = 4π(25)(dr/dt) → dr/dt = 50/(100π) = 1/(2π) cm/s