Differentiation rules allow you to find derivatives of composite, product, quotient, and power functions without returning to first principles every time.
## Core concept
A differentiation rule is a shortcut formula that tells you how to find the derivative (rate of change) of a function based on its structure. Rather than using the limit definition repeatedly, these rules let you quickly compute derivatives for most functions you'll encounter.
Why rules matter: Exam questions ask you to differentiate functions in 1–2 minutes. The limit definition would take 10+ minutes per problem.
## Formula / rule
Power Rule (foundation of most rules) - If $f(x) = x^n$, then $f'(x) = nx^{n-1}$ - Works for any real exponent $n$
Constant Rule - If $f(x) = k$ (constant), then $f'(x) = 0$
Constant Multiple Rule - If $f(x) = k \cdot u(x)$, then $f'(x) = k \cdot u'(x)$
Sum/Difference Rule - $\frac{d}{dx}[u(x) \pm v(x)] = u'(x) \pm v'(x)$ - Differentiate term-by-term
Product Rule - If $f(x) = u(x) \cdot v(x)$, then: $$f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)$$ - Mnemonic: "First × derivative of second + Second × derivative of first"
Quotient Rule - If $f(x) = \frac{u(x)}{v(x)}$, then: $$f'(x) = \frac{u'(x) \cdot v(x) - u(x) \cdot v'(x)}{[v(x)]^2}$$ - Mnemonic: "Low d-high minus high d-low, square the bottom and away we go"
Chain Rule (composite functions) - If $f(x) = g(h(x))$, then: $$f'(x) = g'(h(x)) \cdot h'(x)$$ - Differentiate the outer function, multiply by derivative of the inner function
## Common exam applications
- Polynomial differentiation: Combine power rule + sum rule
- - $f(x) = 3x^4 - 2x^2 + 5 \Rightarrow f'(x) = 12x^3 - 4x$
2. Optimization problems: Use product/quotient rule to find critical points, then apply Maxima/Minima rules
3. Cost/Revenue functions: Marginal cost = $\frac{dC}{dx}$; marginal revenue = $\frac{dR}{dx}$ - Requires quotient rule if cost involves fractions
4. Rate of change: Any function involving time requires chain rule if the variable is composite
## Worked example
Find the derivative of $f(x) = (2x^3 + 1)(x^2 - 3)$
Using product rule: Let $u = 2x^3 + 1$ and $v = x^2 - 3$ - $u' = 6x^2$ - $v' = 2x$ - $f'(x) = 6x^2(x^2 - 3) + (2x^3 + 1)(2x)$ - $f'(x) = 6x^4 - 18x^2 + 4x^4 + 2x$ - $f'(x) = 10x^4 - 18x^2 + 2x$
## Common mistakes
- Forgetting the chain rule: $\frac{d}{dx}(x^2 + 1)^3 \neq 3(x^2 + 1)^2$ — you must multiply by $\frac{d}{dx}(x^2 + 1) = 2x$
- Misapplying quotient rule: Writing $\frac{u' \cdot v'}{v^2}$ instead of the correct $\frac{u'v - uv'}{v^2}$
- Sign errors in quotient rule: The minus sign in the numerator is critical
- Applying product rule to sums: Use sum rule for $u + v$, product rule only for $u \times v$
- Forgetting constants: $\frac{d}{dx}(5x^3) = 15x^2$, not $x^2$
Always simplify fully and check dimensions (if $x$ is quantity, derivative should be rate per unit).