Difficulty: Intermediate | Prerequisites: Limits and Continuity notes, exponent rules
Differentiation is the central skill in a business calculus course. Once you can take derivatives reliably, you unlock marginal analysis, optimisation, and curve sketching. This unit covers the mechanical rules for finding dy/dx: power rule, product rule, quotient rule, chain rule, and the derivatives of exponential, logarithmic, and square-root functions. If your exponent laws or fraction algebra is shaky, revisit that first.
Derivatives measure instantaneous rate of change. The power rule handles most polynomial terms. The chain rule is required whenever a function is nested inside another function. Product and quotient rules handle multiplication and division of two non-trivial expressions.
Derivative (dy/dx or f'(x))
The instantaneous rate of change of a function at a point. Geometrically, it is the slope of the tangent line. Think of it as: "How fast is y changing right now, at this exact value of x?"
Power rule
If f(x) = xⁿ, then f'(x) = n·xⁿ⁻¹. Works for any real exponent, including negatives and fractions.
Product rule
For y = u·v: y' = u'v + uv'. In simple terms, this means you differentiate each factor in turn while keeping the other one untouched, then add the two results.
Chain rule
For y = f(g(x)): dy/dx = f'(g(x))·g'(x). Think of it as: "Derivative of the outside times the derivative of the inside."
Marginal cost / marginal revenue / marginal profit
The derivative of the cost, revenue, or profit function. It estimates the change in cost, revenue, or profit from producing one additional unit.
Percentage rate of change
[f'(x) / f(x)] × 100. Tells you by what percentage the function is changing at a given x.
d/dx [xⁿ] = n·xⁿ⁻¹
This covers negative and fractional exponents too. Rewrite roots and reciprocals as powers first.
√x = x^(1/2), so d/dx [√x] = (1/2)x^(-1/2)
1/x = x⁻¹, so d/dx [1/x] = -x⁻²
1/x³ = x⁻³, so d/dx [1/x³] = -3x⁻⁴
d/dx [c·f(x)] = c·f'(x)
d/dx [f(x) ± g(x)] = f'(x) ± g'(x)
A recurring pattern on the exam: simplify the expression into individual power terms before applying the power rule.
Example: y = (x³ - 3x²)/(2x). Rewrite as x³·(1/2)x⁻¹ - 3x²·(1/2)x⁻¹ = (1/2)x² - (3/2)x. Then y' = x - 3/2. Much easier than the quotient rule here.
Example: y = 1/(3x³) + √x/x. Rewrite as (1/3)x⁻³ + x^(1/2)·x⁻¹ = (1/3)x⁻³ + x^(-1/2). Then y' = -x⁻⁴ - (1/2)x^(-3/2).
Used whenever you have a "function inside a function" (a composition).
Formula: if y = f(g(x)), then y' = f'(g(x)) · g'(x).
Identify the outer function and the inner function. Differentiate the outer (leaving the inner untouched), then multiply by the derivative of the inner.
Example: y = √(x³ + (1/3)x⁻³). Outer: ( )^(1/2). Inner: x³ + (1/3)x⁻³. y' = (1/2)(x³ + (1/3)x⁻³)^(-1/2) · (3x² - x⁻⁴).
Example: y = √(4x + (2/x)). Rewrite as (4x + 2x⁻¹)^(1/2). y' = (1/2)(4x + 2x⁻¹)^(-1/2) · (4 - 2x⁻²).
Example: y = (2x² + 3x)⁻¹. Chain rule gives y' = -1·(2x² + 3x)⁻² · (4x + 3).
Formula: if y = u · v, then y' = u'v + uv'.
Example: y = (x² + 1)³ · (5x³ + 3)³. Let u = (x² + 1)³ and v = (5x³ + 3)³. u' = 3(x² + 1)² · 2x = 6x(x² + 1)². v' = 3(5x³ + 3)² · 15x = 45x(5x³ + 3)². Then y' = 6x(x² + 1)²(5x³ + 3)³ + (x² + 1)³ · 45x(5x³ + 3)².
Many exam problems combine the chain rule with the product rule.
Example: y = 2w² + 3w + 1 where w = x/(x+1) = x(x+1)⁻¹. First find y' in terms of w: y'(w) = 4w + 3. Then find w' using the product rule (or quotient rule): w' = (x+1)⁻². Finally, dy/dx = y'(w) · w' = (4(x+1)⁻¹ + 3)(x+1)⁻².
The exam instruction "do not simplify" means leave your answer in factored or unsimplified form. Do not waste time combining terms.
d/dx [eˣ] = eˣ
d/dx [e^(g(x))] = e^(g(x)) · g'(x) (chain rule)
Example: y = e^(x⁻¹). Then y' = e^(x⁻¹) · (-x⁻²).
d/dx [ln(x)] = 1/x
d/dx [ln(g(x))] = g'(x)/g(x) (chain rule)
Example: d/dx [ln(x³+1)] = 3x²/(x³+1).
Example from the course: y = e^(1/x) · ln(x³+1). Product rule: y' = e^(1/x)·(-x⁻²)·ln(x³+1) + e^(1/x)·[3x²/(x³+1)]. Simplified: e^(1/x)[(-x⁻²)ln(x³+1) + 3x²/(x³+1)].
Always rewrite √( ) as ( )^(1/2) before differentiating.
y = √w where w = x(x+1)⁻¹. Then y = w^(1/2), so y' = (1/2)w^(-1/2) · w'.
Students often forget the chain rule when differentiating composite functions. If there is anything other than a plain x inside a function, the chain rule is needed. Every time.
Students sometimes apply the product rule when they should simplify first. If a fraction can be split into separate power terms (divide each term in the numerator by the denominator), that is usually faster and less error-prone.
When the exam says "do not simplify," students waste time combining terms. Leave the answer in its expanded product/chain form. The marks are for the derivative, not for the algebra.
Students sometimes confuse the product rule and the chain rule. Product rule: two functions multiplied together at the same level. Chain rule: one function inside another (nested).
⚠️ Test II is entirely "find dy/dx, do not simplify." Speed and accuracy with the rules are what matter.
⚠️ The chain rule appears in nearly every problem. If you are unsure whether to use it, use it. You will almost never be penalised for an unnecessary chain rule step (d/dx of x is just 1).
⚠️ Rewriting expressions before differentiating (turning fractions into negative exponents, roots into fractional exponents) is tested directly.
⚠️ Product rule combined with chain rule is a favourite exam pattern: y = (something)ⁿ · (something else)ⁿ.
⚠️ Exponential and logarithmic derivatives will reappear in the applied problems (marginal cost, population growth).
True or false: d/dx [x⁻³] = -3x⁻⁴.
True.
Fill in the blank: The product rule states y' = ______ + ______.
u'v + uv'.
True or false: d/dx [ln(5x)] = 1/(5x).
False. By chain rule, d/dx [ln(5x)] = 5/(5x) = 1/x. (Though the answer happens to be the same as d/dx [ln(x)], the reasoning matters.)
Fill in the blank: To differentiate y = (3x+1)⁵, apply the ______ rule.
Chain rule.
True or false: d/dx [e^(2x)] = e^(2x).
False. d/dx [e^(2x)] = 2e^(2x). The chain rule requires multiplying by the derivative of 2x.
Q: Find dy/dx for y = 1/(3x³) + √x/x. Do not simplify.
A: Rewrite as (1/3)x⁻³ + x^(-1/2). Then y' = -x⁻⁴ - (1/2)x^(-3/2).
Q: Find dy/dx for y = (x² + 1)³(5x³ + 3)³.
A: Product rule. u = (x²+1)³, u' = 6x(x²+1)². v = (5x³+3)³, v' = 45x²(5x³+3)². y' = 6x(x²+1)²(5x³+3)³ + (x²+1)³·45x²(5x³+3)².
Q: Find dy/dx for y = √(4x + 2/x).
A: Rewrite as (4x + 2x⁻¹)^(1/2). Chain rule: y' = (1/2)(4x + 2x⁻¹)^(-1/2) · (4 - 2x⁻²).
Q: Find dy/dx for y = e^(1/x) · ln(x³+1).
A: Product rule with chain rule. y' = e^(1/x)·(-x⁻²)·ln(x³+1) + e^(1/x)·[3x²/(x³+1)].
Q: Find dy/dx for y = 1/(2x²+3x).
A: Rewrite as (2x²+3x)⁻¹. Chain rule: y' = -1·(2x²+3x)⁻² · (4x+3).
These derivative rules are the tools you use for every applied topic that follows. Marginal analysis (next unit) is just "take the derivative of the cost or revenue function." Optimisation requires setting the derivative to zero to find critical points. Curve sketching uses both the first derivative (increasing/decreasing) and the second derivative (concavity).
Related Terms / Search Tags
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