Difficulty: Intermediate | Prerequisites: Limits, basic algebra, exponent rules, logarithm properties.
Derivatives are the central skill in MAC 2233. Every applied problem in this course, from marginal cost to profit maximisation, requires you to differentiate something. This set of notes covers the mechanics: the rules you need to find derivatives quickly, the formal definition that underpins them, and how to handle exponentials, logarithms, and compositions. If you can differentiate confidently, the application sections become straightforward. If you cannot, they become guesswork.
TL;DR: The derivative measures instantaneous rate of change. You compute it using rules (power, product, chain) or, when required, the limit definition. The chain rule handles compositions, the product rule handles products, and exponentials and logarithms have their own patterns.
Derivative
The instantaneous rate of change of a function at a point. Written f'(x) or dy/dx. Think of it as the slope of the curve at a single point, rather than the slope between two points.
Power rule
If f(x) = xⁿ, then f'(x) = nxⁿ⁻¹. In simple terms, bring the exponent down as a coefficient, then reduce the exponent by one.
Product rule
For y = u(x) · v(x), the derivative is y' = u'v + uv'. Think of it as: differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second.
Chain rule
For y = f(g(x)), the derivative is 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.
Formal definition of the derivative
f'(x) = lim (h→0) [f(x+h) − f(x)] / h. This is the limit process. Think of it as shrinking the secant line down to a tangent line by making the gap between the two points vanishingly small.
Marginal average cost
The derivative of the average cost function. Average cost is C̄(x) = C(x)/x. The marginal average cost, C̄'(x), tells you how the per-unit cost changes as you produce one more unit.
Natural logarithm derivative
If y = ln(x), then y' = 1/x. For y = ln(g(x)), by the chain rule, y' = g'(x)/g(x).
Exponential derivative
If y = eˣ, then y' = eˣ. For y = e^(g(x)), by the chain rule, y' = g'(x) · e^(g(x)).
If y = axⁿ, then y' = a · n · xⁿ⁻¹. Constants vanish (derivative of a constant is 0).
Example: y = 7x⁵ + 10x − 8. Derivative: y' = 35x⁴ + 10. The constant −8 drops out.
For y = u · v, compute y' = u'v + uv'.
Example: y = x³ · e²ˣ. Let u = x³, v = e²ˣ. Then u' = 3x², v' = 2e²ˣ. So y' = 3x² · e²ˣ + x³ · 2e²ˣ = x²e²ˣ(3 + 2x) after factoring.
The factored form is what exams typically expect. Always look for common factors to pull out.
For y = f(g(x)), differentiate the outer function evaluated at the inner, then multiply by the derivative of the inner.
Example: y = 4(ln x)⁶. Outer function: u⁶. Inner function: ln x. Derivative of outer: 6u⁵ = 6(ln x)⁵. Derivative of inner: 1/x. Result: y' = 4 · 6(ln x)⁵ · (1/x) = 24(ln x)⁵ / x.
Example: y = e^(x² + 5x + 4). Outer: eᵘ. Inner: x² + 5x + 4. Derivative: (2x + 5) · e^(x² + 5x + 4).
Example: y = 3x(2x² + 5)³. Use the product rule with u = 3x and v = (2x² + 5)³.
u' = 3.
v' = 3(2x² + 5)² · 4x = 12x(2x² + 5)² (chain rule on the cubic).
y' = 3(2x² + 5)³ + 3x · 12x(2x² + 5)² = 3(2x² + 5)²[(2x² + 5) + 12x²] = 3(2x² + 5)²(14x² + 5).
f'(x) = lim (h→0) [f(x+h) − f(x)] / h.
Example: f(x) = x².
f(x+h) = (x+h)² = x² + 2xh + h².
f(x+h) − f(x) = 2xh + h².
Divide by h: 2x + h.
Take the limit as h→0: f'(x) = 2x.
This method is slow but exams may require it explicitly. The power rule gives the same answer instantly.
Start with total cost C(x). Average cost: C̄(x) = C(x)/x.
Marginal average cost: differentiate C̄(x) using the quotient rule (or simplify first, then use the power rule).
Example: C(x) = 4x² + 2x. Average cost: C̄(x) = 4x + 2. Marginal average cost: C̄'(x) = 4.
If C̄(x) = 4x + 2/x, then C̄'(x) = 4 − 2/x² (using the power rule on 2x⁻¹).
Example: f(x) = 5x⁷ ln x. Product rule: u = 5x⁷, v = ln x.
u' = 35x⁶, v' = 1/x.
f'(x) = 35x⁶ · ln x + 5x⁷ · (1/x) = 35x⁶ ln x + 5x⁶ = 5x⁶(7 ln x + 1).
\frac{d}{dx}[x^n] = nx^{n-1}\frac{d}{dx}[u \cdot v] = u'v + uv'\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\frac{d}{dx}[e^{g(x)}] = g'(x) \cdot e^{g(x)}\frac{d}{dx}[\ln(g(x))] = \frac{g'(x)}{g(x)}Derivatives are how businesses compute marginal cost, marginal revenue, and marginal profit: the additional cost or revenue from producing one more unit. In finance, the derivative of a growth function gives you the rate of growth at any point, which is essential for modelling interest, depreciation, and investment returns.
Students often forget to multiply by the inner derivative when using the chain rule. The chain rule always has two parts: the outer derivative and the inner derivative.
When using the product rule, students sometimes differentiate both factors simultaneously (u' · v') instead of applying u'v + uv'. The product rule is a sum, not a product of derivatives.
Students sometimes drop constants when differentiating. The derivative of a constant term is zero, but a constant multiplied by a variable term stays as a coefficient.
With the formal definition, students forget to cancel h from the numerator before taking the limit, which leaves 0/0. You must simplify algebraically first.
⚠️ The formal definition (limit process) question will award zero credit if you use shortcut rules. Show every algebraic step.
⚠️ Product-and-chain combined problems (like y = 3x(2x² + 5)³) are high-mark questions. Factor completely after differentiating.
⚠️ Forgetting the chain rule on e^(g(x)) is one of the most common errors on this exam. The derivative is g'(x) · e^(g(x)), not just e^(g(x)).
⚠️ On log-times-power problems (like 5x⁷ ln x), you must use the product rule. Factor the result to earn full marks.
Fill in the blank: The derivative of 7x⁵ is ___.
35x⁴.
True or False: The derivative of e^(3x) is e^(3x).
False. It is 3e^(3x) (chain rule).
Fill in the blank: The product rule states (uv)' = ___.
u'v + uv'.
True or False: In the formal definition, you can substitute h = 0 directly.
False. You must simplify first, then take the limit.
Q: Find the derivative of y = 7x⁵ + 10x − 8.
A: y' = 35x⁴ + 10. The constant −8 disappears. The 10x becomes 10.
Q: Differentiate y = 4(ln x)⁶ using the chain rule.
A: y' = 24(ln x)⁵ / x. Outer derivative: 4 · 6(ln x)⁵. Inner derivative: 1/x.
Q: Calculate the derivative of y = x³e²ˣ and factor completely.
A: Product rule gives 3x²e²ˣ + 2x³e²ˣ. Factor out x²e²ˣ to get x²e²ˣ(3 + 2x).
Q: Find the derivative of y = e^(x² + 5x + 4).
A: By the chain rule, y' = (2x + 5)e^(x² + 5x + 4).
Q: Use the formal definition to find f'(x) for f(x) = x².
A: f'(x) = lim (h→0) [(x+h)² − x²] / h = lim (h→0) [2xh + h²] / h = lim (h→0) (2x + h) = 2x.
Q: Differentiate y = 3x(2x² + 5)³.
A: y' = 3(2x² + 5)²(14x² + 5). Product rule plus chain rule, then factor.
Q: Differentiate f(x) = 5x⁷ ln x and factor.
A: f'(x) = 35x⁶ ln x + 5x⁶ = 5x⁶(7 ln x + 1).
Derivatives connect directly to tangent lines (the derivative gives the slope), to optimisation (set the derivative to zero to find critical points), and to marginal analysis (the derivative of cost, revenue, or profit). The formal definition circles back to limits, reinforcing that calculus is built on the limit concept.
derivative, differentiation, power rule, product rule, chain rule, formal definition of derivative, limit definition, exponential derivative, logarithmic derivative, marginal average cost, e to the x, ln x, composition of functions, MAC 2233, calculus for business, FIU