IP AMaths Notes (Upper Sec, Year 3-4): 14) Differentiation Fundamentals

Differentiation measures instantaneous rate of change. Know how to move from first principles to the standard rules.

Key rules

Limit definition. \(f^{\prime}(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h}\).

Power rule (integer (n )). \(\dfrac{\mathrm{d}}{\mathrm{d}x}\bigl(x^n\bigr) = nx^{n - 1}\).

Constant multiple. \(\dfrac{\mathrm{d}}{\mathrm{d}x}[c \cdot f(x)] = c \cdot f^{\prime}(x)\).

Sum or difference. \(\dfrac{\mathrm{d}}{\mathrm{d}x}[f(x) \pm g(x)] = f^{\prime}(x) \pm g^{\prime}(x)\).

Worked example 1 — From first principles

Find \( f^{\prime}(x) \) for \( f(x) = x^2 \) using the limit definition.

1. Compute difference quotient: \( \dfrac{(x + h)^2 - x^2}{h} = \dfrac{x^2 + 2xh + h^2 - x^2}{h} \).
2. Simplify: \( \dfrac{2xh + h^2}{h} = 2x + h \).
3. Take limit: as \( h \to 0 \), \( f^{\prime}(x) = 2x \).

Worked example 2 — Composite function

Find \( \dfrac{\mathrm{d}y}{\mathrm{d}x} \) for \( y = 5x^4 - 3x^2 + 7 \).

1. Differentiate term by term: \( \dfrac{\mathrm{d}}{\mathrm{d}x}(5x^4) = 20x^3 \).
2. \( \dfrac{\mathrm{d}}{\mathrm{d}x}(-3x^2) = -6x \); constant term differentiates to \( 0 \).
3. Combine: \( \dfrac{\mathrm{d}y}{\mathrm{d}x} = 20x^3 - 6x \).

Try this

Differentiate \( y = \dfrac{3}{\sqrt{x}} + \dfrac{2}{x^2} \) and simplify the result.