Difficulty: Intermediate | Prerequisites: Limits and Continuity notes, exponent and logarithm laws
Differentiation is the machinery for finding how fast a function is changing at any given point. In a business calculus course, this is the tool behind every "rate of change," "marginal" analysis and optimisation problem you will meet. You need the limit definition to understand what a derivative is, but the rules in these notes are how you compute one efficiently. If you are comfortable with limits and continuity, you are ready for this material.
The derivative measures the instantaneous rate of change of a function. A handful of rules (power, product, quotient, chain) let you differentiate virtually any function you will see on this exam. Exponential and logarithmic derivatives have their own patterns, and logarithmic differentiation handles complicated products, quotients and powers in one technique.
Derivative
The instantaneous rate of change of f(x) at a point. Formally, f'(x) = lim(h→0) [f(x+h) − f(x)] / h. In simple terms, it is the slope of the tangent line at any point on the curve.
Power rule
If f(x) = x^n, then f'(x) = nx^(n−1). This covers integer, fractional and negative exponents. Think of it as: bring the exponent down as a coefficient and reduce it by one.
Product rule
If y = F · S (two functions multiplied), then y' = F'S + FS'. In simple terms, differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second.
Quotient rule
If y = T/B (top over bottom), then y' = (T'B − TB') / B². Think of it as: "low d-high minus high d-low, over low squared."
Chain rule
If y = f(g(x)), then y' = f'(g(x)) · g'(x). In simple terms, differentiate the outer function (leaving the inner function alone), then multiply by the derivative of the inner function.
Logarithmic differentiation
A technique for differentiating products, quotients and powers that are too messy for ordinary rules. Take ln of both sides, use log laws to simplify, differentiate implicitly, then multiply both sides by the original function.
Marginal cost / revenue / profit
The derivative of the cost, revenue or profit function. It approximates the change in cost, revenue or profit from producing one additional unit. In simple terms, it answers "how much more does the next unit cost (or earn)?"
The formal (limit) definition: f'(x) = lim(h→0) [f(x+h) − f(x)] / h. The exam may require you to use this four-step process, with no shortcut rules, for a specific function.
Example: f(x) = 5/x.
f(x+h) = 5/(x+h).
f(x+h) − f(x) = 5/(x+h) − 5/x = [5x − 5(x+h)] / [x(x+h)] = −5h / [x(x+h)].
Divide by h: −5 / [x(x+h)].
Take lim(h→0): −5/x².
Rewrite roots and reciprocals as fractional/negative exponents before applying the rule.
√x = x^(1/2), so d/dx[√x] = (1/2)x^(−1/2) = 1/(2√x).
1/x² = x^(−2), so d/dx[1/x²] = −2x^(−3) = −2/x³.
Example: y = 3√x − 7x⁵ + 10x − 8. Rewrite: 3x^(1/2) − 7x⁵ + 10x − 8. Then y' = (3/2)x^(−1/2) − 35x⁴ + 10 = 3/(2√x) − 35x⁴ + 10.
(FS)' = F'S + FS'. Keep both original functions and differentiate one at a time.
Example: y = (3x² − 5)(x³ + 1). F = 3x²−5, F' = 6x. S = x³+1, S' = 3x².
y' = 6x(x³+1) + (3x²−5)(3x²) = 6x⁴+6x + 9x⁴−15x² = 15x⁴ − 15x² + 6x.
Example: f(x) = 5x⁷ ln x. F = 5x⁷, S = ln x. f'(x) = 35x⁶ ln x + 5x⁷(1/x) = 35x⁶ ln x + 5x⁶ = 5x⁶(1 + 7 ln x).
y' = (T'B − TB') / B². Common trick: simplify first if possible.
Example: y = (5x² − 7x)/x. Simplify to 5x − 7 before differentiating. y' = 5. Much easier than applying the quotient rule.
Example: y = e^(4x) / (2x−1)³. Quotient rule gives y' = [4e^(4x)(2x−1)³ − e^(4x) · 3(2x−1)² · 2] / (2x−1)⁶. Factor: e^(4x)(2x−1)²[4(2x−1) − 6] / (2x−1)⁶ = 2e^(4x)(4x−5) / (2x−1)⁴.
Differentiate the outer function, leave the inner alone, then multiply by the inner's derivative.
Example: y = e^(x²−5x+4). Outer: e^(something). Inner: x²−5x+4. y' = e^(x²−5x+4) · (2x−5).
Example: y = 3x(2x²−5)³. Product rule + chain rule. y' = 3(2x²−5)³ + 3x · 3(2x²−5)² · 4x = 3(2x²−5)³ + 36x²(2x²−5)². Factor out 3(2x²−5)²: 3(2x²−5)²[(2x²−5) + 12x²] = 3(2x²−5)²(14x²−5).
d/dx[e^x] = e^x.
d/dx[e^(u)] = e^(u) · u' (chain rule).
d/dx[ln x] = 1/x.
d/dx[ln u] = u'/u (chain rule).
d/dx[ln(xⁿ)] = n/x (bring the exponent down using log laws first).
Examples:
y = 4 ln x⁶. Rewrite as 24 ln x. Then y' = 24/x.
y = ln(3x²/e^(2x)). Use log laws: ln(3x²) − 2x = ln 3 + 2 ln x − 2x. Then y' = 2/x − 2.
Useful when the function is a complex product/quotient with variable exponents or roots.
Example: y = x⁵ · ³√(x²−1) / e^(3x).
Take ln: ln y = 5 ln x + (1/3) ln(x²−1) − 3x.
Differentiate: y'/y = 5/x + (2x)/(3(x²−1)) − 3.
Multiply both sides by y: y' = [5/x + 2x/(3(x²−1)) − 3] · [x⁵³√(x²−1) / e^(3x)].
Differentiate the first derivative again.
Example: y = e^(3x²). First derivative: y' = 6x · e^(3x²). Second derivative (product rule + chain rule): y'' = 6e^(3x²) + 6x · 6x · e^(3x²) = 6e^(3x²)(1 + 6x²).
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\frac{d}{dx}[x^n] = nx^{n-1}(FS)' = F'S + FS' \quad \text{(product rule)}\left(\frac{T}{B}\right)' = \frac{T'B - TB'}{B^2} \quad \text{(quotient rule)}\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \quad \text{(chain rule)}\frac{d}{dx}[e^u] = e^u \cdot u', \qquad \frac{d}{dx}[\ln u] = \frac{u'}{u}The derivative of a cost function gives the marginal cost, which tells a business how much extra it costs to produce one more item. Revenue derivatives reveal when adding another unit of sales starts to bring diminishing returns. These ideas underpin pricing, production and inventory decisions in any firm.
Students write the product rule as (FS)' = F'S − FS'. The product rule uses addition, not subtraction. The quotient rule is the one with the minus sign.
When differentiating ln(3x), students often write 3/x. The correct answer is 1/x: the chain rule gives (1/(3x)) · 3 = 1/x. (Note: ln(3x) = ln 3 + ln x, and the derivative of the constant ln 3 is zero.)
Students forget to multiply by the inner derivative when using the chain rule. For y = e^(x²), the derivative is 2x · e^(x²), not just e^(x²).
Profit = Revenue − Cost, not Cost − Revenue. Getting this backwards flips the sign of every answer.
⚠️ The definition-of-derivative problem (Q25) requires the limit process. Using shortcut rules earns zero credit.
⚠️ "Factorize completely" appears on several problems. After differentiating, you must factor out common terms. Partial marks are lost for unfactored answers.
⚠️ Logarithmic differentiation (Q37) is tested. Practise the full process: take ln, simplify, differentiate implicitly, multiply by y.
⚠️ Know when to simplify before differentiating. If a fraction can be reduced (like (5x²−7x)/x = 5x−7), simplifying first avoids unnecessary quotient-rule work.
True or false: The derivative of e^(5x) is e^(5x). (False. By the chain rule it is 5e^(5x).)
Fill in the blank: d/dx[ln x⁶] = ______/x. (6.)
True or false: The product rule is (FS)' = F'S'. (False. It is F'S + FS'.)
Fill in the blank: To differentiate y = x^(−3), bring the −3 down and reduce the exponent to get y' = ______. (−3x^(−4).)
Q: Differentiate y = 3√x − 7x⁵ + 10x − 8.
A: Rewrite as 3x^(1/2) − 7x⁵ + 10x − 8. Apply the power rule term by term: y' = (3/2)x^(−1/2) − 35x⁴ + 10 = 3/(2√x) − 35x⁴ + 10.
Q: Differentiate y = x³e^(2x). Factor completely.
A: Product rule. F = x³, F' = 3x². S = e^(2x), S' = 2e^(2x). y' = 3x²e^(2x) + 2x³e^(2x) = x²e^(2x)(3 + 2x).
Q: Differentiate y = e^(x²−5x+4).
A: Chain rule. Outer: e^(u). Inner: x²−5x+4, inner derivative: 2x−5. y' = (2x−5)e^(x²−5x+4).
Q: Differentiate y = 3x(2x²−5)³. Factorize completely.
A: Product + chain rule. y' = 3(2x²−5)³ + 3x · 3(2x²−5)² · 4x = 3(2x²−5)³ + 36x²(2x²−5)². Factor out 3(2x²−5)²: y' = 3(2x²−5)²(14x²−5).
Q: Differentiate y = 4 ln x⁶.
A: Rewrite using log laws: y = 24 ln x. Then y' = 24/x.
Q: Differentiate y = (3x²−5)(x³+1).
A: Product rule. y' = 6x(x³+1) + (3x²−5)(3x²) = 6x⁴ + 6x + 9x⁴ − 15x² = 15x⁴ − 15x² + 6x.
Q: Differentiate f(x) = 5x⁷ ln x. Factorize completely.
A: Product rule. f'(x) = 35x⁶ ln x + 5x⁷ · (1/x) = 35x⁶ ln x + 5x⁶ = 5x⁶(1 + 7 ln x).
Q: Differentiate y = ln(3x²/e^(2x)).
A: Simplify first. y = ln(3x²) − ln(e^(2x)) = ln 3 + 2 ln x − 2x. Then y' = 2/x − 2.
Q: Find the second derivative of y = e^(3x²). Factorize completely.
A: y' = 6xe^(3x²). Then y'' (product + chain rule) = 6e^(3x²) + 6x · 6xe^(3x²) = 6e^(3x²)(1 + 6x²).
Q: Use the definition of the derivative to find f'(x) for f(x) = 5/x.
A: f(x+h) = 5/(x+h). [f(x+h)−f(x)]/h = [5/(x+h) − 5/x]/h = −5h/[hx(x+h)] = −5/[x(x+h)]. Take lim(h→0): −5/x².
The derivative is defined using limits (covered in the Limits and Continuity notes). Derivatives of cost, revenue and profit functions lead directly into the business applications and optimisation problems covered in the Applications of Derivatives notes. The second derivative connects to concavity and inflection points, which you need for curve sketching. Integration (the next major topic) reverses differentiation, so every rule here has a mirror image in the antiderivative rules.
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