Difficulty: Intermediate. Prerequisites: comfort with limits, basic algebra including exponent rules and logarithm properties.
Differentiation is the process of finding the derivative, which tells you the instantaneous rate of change of a function. This course covers the power rule, product rule, quotient rule, chain rule, and specialised rules for exponentials and logarithms. Most exam problems combine two or more of these rules in a single question, so fluency with each rule individually is essential before tackling combinations.
Derivative (f'(x) or dy/dx)
The instantaneous rate of change of f(x) with respect to x. Geometrically, it is the slope of the tangent line at a point. Think of it as: "How fast is y changing right now, at this exact x-value?"
Power rule
If f(x) = x^n, then f'(x) = n * x^(n-1). Works for any real exponent, including fractions and negatives. In simple terms: bring the exponent down as a coefficient, then reduce the exponent by one.
Product rule
If y = f(x) * g(x), then y' = f'(x) * g(x) + f(x) * g'(x). Think of it as: "derivative of the first times the second, plus the first times the derivative of the second."
Quotient rule
If y = f(x) / g(x), then y' = (f'(x) * g(x) - f(x) * g'(x)) / [g(x)]^2. In simple terms: "low d-high minus high d-low, all over the square of what's below."
Chain rule
If y = f(g(x)), then y' = f'(g(x)) * g'(x). Differentiate the outer function at the inner function, then multiply by the derivative of the inner function. Think of it as: peeling layers. Differentiate outside-in, multiplying each layer's derivative.
Exponential differentiation
The derivative of e^x is e^x. For e^(u(x)), use the chain rule: derivative is e^(u(x)) * u'(x). In simple terms: e^(something) stays as it is, then multiply by the derivative of the something.
Logarithmic differentiation
The derivative of ln(x) is 1/x. For ln(u(x)), chain rule gives u'(x) / u(x). Logarithmic differentiation is a technique where you take ln of both sides of a complicated product/quotient/power expression, differentiate implicitly, then solve for y'. Think of it as: a workaround for messy products and quotients that would be painful with the product/quotient rule alone.
Marginal average cost
The derivative of the average cost function C-bar(x) = C(x)/x. It tells you how the average cost per unit changes when you produce one more unit. In simple terms: the rate at which your per-unit cost is rising or falling as production changes.
Rewrite radicals and denominators as fractional/negative exponents before differentiating.
Example (Q3): y = 3sqrt(x) - 7x^5 + 10x - 8. Rewrite sqrt(x) as x^(1/2). Then y' = 3 * (1/2) * x^(-1/2) - 35x^4 + 10 = (3)/(2sqrt(x)) - 35x^4 + 10. Answer: A.
Constants differentiate to zero. The derivative of -8 is 0.
The derivative of ln(x^6): use the log power rule to write 6*ln(x), then differentiate to get 6/x. Alternatively, chain rule on ln(x^6) gives 6x^5 / x^6 = 6/x. Both paths reach the same answer.
Q9: y = 4ln(x^6) = 24ln(x), so y' = 24/x. Answer: D.
Apply when two functions are multiplied together.
Q10: y = x^3 * e^(2x). Let f = x^3, g = e^(2x). Then f' = 3x^2, g' = 2e^(2x). Product rule: y' = 3x^2 * e^(2x) + x^3 * 2e^(2x) = x^2 * e^(2x) * (3 + 2x). Answer: D.
Q33a: y = (3x^2 - 5)(x^3 + 1). Product rule gives y' = 6x(x^3 + 1) + (3x^2 - 5)(3x^2) = 6x^4 + 6x + 9x^4 - 15x^2 = 15x^4 - 15x^2 + 6x.
Important identity from Q11: (f * g)' = f' * g + f * g' (not f' * g - f * g'). Statement I in Q11 is wrong because it states (FS)' = F'S - FS'.
Apply when one function is divided by another.
Q8: Average cost C-bar(x) = C(x)/x = (4x + 8000)/x = 4 + 8000/x = 4 + 8000*x^(-1). Marginal average cost: C-bar'(x) = -8000/x^2. Answer: B.
Q27b: y = (5x^2 - 7x)/x. Simplify first to 5x - 7, then y' = 5. Simplifying before differentiating saves time.
Used whenever you differentiate a function of a function (a composition).
Q12: y = e^(x^2 - 5x + 4). Outer function is e^u, inner is u = x^2 - 5x + 4. Chain rule: y' = e^(x^2 - 5x + 4) * (2x - 5). Answer: D.
Q13: y = 3x(2x^2 - 5)^3. This needs both the product rule and the chain rule. Let f = 3x, g = (2x^2 - 5)^3. Then f' = 3, g' = 3(2x^2 - 5)^2 * 4x = 12x(2x^2 - 5)^2. Product rule: y' = 3(2x^2 - 5)^3 + 3x * 12x(2x^2 - 5)^2 = 3(2x^2 - 5)^2 * [(2x^2 - 5) + 12x^2] = 3(2x^2 - 5)^2 * (14x^2 - 5). Answer: A.
d/dx [e^x] = e^x
d/dx [e^(u)] = e^(u) * u'
d/dx [ln(x)] = 1/x
d/dx [ln(u)] = u'/u
Q33b: f(x) = 5x^7 * ln(x). Product rule: f'(x) = 35x^6 * ln(x) + 5x^7 * (1/x) = 35x^6 * ln(x) + 5x^6 = 5x^6(7*ln(x) + 1).
Q33c: y = ln(3x^2 / e^(2x)). Simplify first using log rules: ln(3x^2) - ln(e^(2x)) = ln(3) + 2*ln(x) - 2x. Then y' = 2/x - 2.
Take the natural log of both sides, simplify with log rules, differentiate implicitly, then multiply by y.
Q37: y = x^5 * (x^2 - 1)^(1/3) / e^(3x). Take ln: ln(y) = 5*ln(x) + (1/3)*ln(x^2 - 1) - 3x. Differentiate: (1/y)*y' = 5/x + (2x)/(3(x^2 - 1)) - 3. Multiply both sides by y to isolate y'.
Q25 asks you to differentiate f(x) = 5/x using the limit definition: f'(x) = lim (h->0) [f(x+h) - f(x)] / h.
Steps: (1) write f(x+h) = 5/(x+h), (2) form [5/(x+h) - 5/x], (3) combine fractions over a common denominator: [5x - 5(x+h)] / [x(x+h)] = -5h / [x(x+h)], (4) divide by h and take the limit: -5 / [x(x+h)] as h->0 gives -5/x^2.
Constant: d/dx [c] = 0
Power rule: d/dx [x^n] = n * x^(n-1)
Constant multiple: d/dx [c * f(x)] = c * f'(x)
Sum/difference: d/dx [f(x) +/- g(x)] = f'(x) +/- g'(x)
Product rule: d/dx [f * g] = f' * g + f * g'
Quotient rule: d/dx [f / g] = (f' * g - f * g') / g^2
Chain rule: d/dx [f(g(x))] = f'(g(x)) * g'(x)
Exponential: d/dx [e^x] = e^x
General exponential chain: d/dx [e^(u)] = e^(u) * u'
Natural log: d/dx [ln(x)] = 1/x
General log chain: d/dx [ln(u)] = u' / u
Profit: Profit = Revenue - Cost (not Cost - Revenue; Q11 statement III is false)
Every marginal quantity in business, whether marginal cost, marginal revenue, or marginal profit, is a derivative. When a company asks "what happens to our costs if we produce one more unit?", the answer is the derivative of the cost function evaluated at the current production level. Logarithmic differentiation appears in modelling compound growth rates, where products and powers of variables make standard rules unwieldy.
Students often forget to apply the chain rule when differentiating e^(something). The derivative of e^(3x) is 3e^(3x), not e^(3x). Always multiply by the derivative of the exponent.
Confusing the product rule and the quotient rule signs. Product rule uses a plus (+). Quotient rule uses a minus (-), and it matters which term comes first in the numerator: f'g - fg', not fg' - f'g.
When a function can be simplified before differentiating (e.g. (5x^2 - 7x)/x = 5x - 7), doing so is faster and less error-prone than applying the quotient rule to the unsimplified form. Always simplify first if you can.
Students sometimes write Profit = Cost - Revenue instead of Profit = Revenue - Cost. This reversal flips the sign of the entire answer.
⚠️ Chain rule combined with product rule is the most common multi-step problem on the exam. Q10, Q13, and Q39 all require this combination.
⚠️ Factoring completely after differentiating is frequently required. If the question says "factor completely," marks are lost for leaving the answer in expanded form.
⚠️ The four-step limit definition of the derivative (Q25) is a classic free-response question. Practise it until the algebra is smooth.
⚠️ Know when to use logarithmic differentiation versus standard rules. If the function has a variable in both the base and the exponent, or is a complex product/quotient of many terms, logarithmic differentiation is the cleaner path.
⚠️ d/dx [ln(x)] = 1/x, and by extension d/dx [ln(3x)] = 1/x (not 1/(3x)). The chain rule gives 3/(3x) = 1/x. This trips students up.
True or False: The derivative of e^(5x) is e^(5x).
False. By the chain rule, d/dx [e^(5x)] = 5e^(5x). You must multiply by the derivative of the exponent.
Fill in the blank: The product rule states (fg)' = f'g ___ fg'.
(plus). The product rule adds the two terms.
True or False: d/dx [ln(x^6)] = 6/x.
True. Either rewrite as 6*ln(x) and differentiate, or apply the chain rule: 6x^5 / x^6 = 6/x.
Fill in the blank: Profit = Revenue ___ Cost.
Minus. Profit = Revenue - Cost.
True or False: To differentiate y = (2x^2 - 5)^3, you only need the power rule.
False. You need the chain rule as well. The power rule alone would miss the inner derivative (4x).
Q: Differentiate y = 3*sqrt(x) - 7x^5 + 10x - 8.
A: Rewrite sqrt(x) as x^(1/2). y' = (3/2)x^(-1/2) - 35x^4 + 10 = 3/(2*sqrt(x)) - 35x^4 + 10.
Q: Differentiate y = x^3 * e^(2x) and factor completely.
A: Product rule with chain rule. y' = 3x^2 * e^(2x) + x^3 * 2e^(2x) = x^2 * e^(2x)(3 + 2x).
Q: Find the marginal average cost given C(x) = 4x + 8,000.
A: Average cost C-bar(x) = (4x + 8000)/x = 4 + 8000/x. C-bar'(x) = -8000/x^2.
Q: Differentiate y = e^(x^2 - 5x + 4).
A: Chain rule. y' = (2x - 5) * e^(x^2 - 5x + 4).
Q: Differentiate y = 3x(2x^2 - 5)^3. Factor completely.
A: Product rule + chain rule. y' = 3(2x^2 - 5)^3 + 3x * 3(2x^2 - 5)^2 * 4x = 3(2x^2 - 5)^2[(2x^2 - 5) + 12x^2] = 3(2x^2 - 5)^2(14x^2 - 5).
Q: Use the limit definition to find f'(x) for f(x) = 5/x.
A: f'(x) = lim (h->0) [5/(x+h) - 5/x] / h = lim (h->0) [-5h / (x(x+h)h)] = lim (h->0) -5/(x(x+h)) = -5/x^2.
Differentiation builds directly on limits (the derivative is defined as a limit) and feeds into every application topic: tangent lines, curve sketching, optimisation, and marginal analysis all require you to differentiate first and then interpret the result.
The chain rule is essential for implicit differentiation and related rates problems if you encounter those in later courses. Logarithmic differentiation connects to exponential growth and decay models used in finance (compound interest, depreciation).
derivative, differentiation, power rule, product rule, quotient rule, chain rule, exponential derivative, e^x, ln(x), natural log derivative, logarithmic differentiation, marginal cost, marginal average cost, factoring derivatives, four-step process, limit definition of derivative, composite function, inner function, outer function, business calculus, MAT101, calculus for business