Difficulty: Beginner to Intermediate | Prerequisites: Chapter 1 (limits and continuity)
This section of Calculus for Business introduces the core rules for taking derivatives: the constant rule, the power rule, and the constant multiple rule. These are the building blocks you will use for every derivative problem in the course, and they sit between the limit definition of the derivative (Chapter 1) and the product rule (covered in the companion study guide). If you can apply these three rules fluently, the more advanced rules are just extensions of the same logic.
Derivatives measure instantaneous rate of change. Three rules handle most basic functions: constants differentiate to zero, powers bring the exponent down and reduce it by one, and constant multiples pass straight through the derivative. Master these and you can differentiate any polynomial or simple rational function by hand.
Derivative, f'(x)
The instantaneous rate of change of a function at a given point. Formally, the limit of the difference quotient as h approaches zero.
In simple terms, it tells you how fast the output is changing at any particular input value.
Constant Rule
The derivative of any constant c is zero: d/dx (c) = 0.
Think of it as: a flat line has no slope, so its rate of change is always zero.
Power Rule
For any real number n, the derivative of x raised to the n is n times x raised to the (n - 1): d/dx (x^n) = n · x^(n-1).
In simple terms, bring the exponent down as a multiplier, then knock the exponent down by one.
Constant Multiple Rule
The derivative of a constant times a function equals the constant times the derivative of that function: d/dx [c · f(x)] = c · f'(x).
Think of it as: constants are along for the ride, they pass straight through the derivative operation.
Negative Exponent
1/x^n can be rewritten as x^(-n). This is not a differentiation rule itself, but an algebraic rewrite you will use constantly to put terms into a form the power rule can handle.
In simple terms, move a power from the denominator to the numerator by flipping the sign of the exponent.
If f(x) = c for some constant c, then f'(x) = 0.
Graphically, a constant function is a horizontal line, and horizontal lines have zero slope everywhere.
If f(x) = x^n for any real number n, then f'(x) = n · x^(n-1).
This works for positive integers, negative integers, and fractional exponents alike.
Example: f(x) = x^3 gives f'(x) = 3x^2.
Example: f(x) = x^(-1) = 1/x gives f'(x) = -1 · x^(-2) = -1/x^2.
Example: f(x) = x^(1/2) = sqrt(x) gives f'(x) = (1/2) · x^(-1/2).
If f(x) = c · g(x), then f'(x) = c · g'(x).
The constant stays put; you only differentiate the function it multiplies.
Example: f(x) = 5x^3 gives f'(x) = 5 · 3x^2 = 15x^2.
Before applying the power rule, rewrite terms so that every part is in the form (constant) · x^(some number).
Fractions with x in the denominator: rewrite 1/x^n as x^(-n).
Example: 1/(3x^4) becomes (1/3) · x^(-4). Then f'(x) = (1/3)(-4)x^(-5) = -4/(3x^5).
Roots: rewrite sqrt(x) as x^(1/2), cube root of x as x^(1/3), and so on.
Combined terms: break a complicated expression into a sum of individual power-rule-friendly terms.
Given f(x) = (-x^4)/16 + (1/2)x^(4/6) - x^(4/6) + 1/(3x^4) + (1/3)x.
Rewrite each term in power-rule form:
(-1/16) · x^4
(1/2) · x^(2/3) (since 4/6 simplifies to 2/3)
-x^(2/3)
(1/3) · x^(-4)
(1/3) · x^1
Differentiate term by term:
f'(x) = (-1/16)(4)x^3 + (1/2)(2/3)x^(-1/3) - (2/3)x^(-1/3) + (1/3)(-4)x^(-5) + 1/3
Simplify: f'(x) = -x^3/4 + (1/3)x^(-1/3) - (2/3)x^(-1/3) - (4/3)x^(-5) + 1/3
The gross domestic product (GDP) of a country was modelled as N(t) = t^2 + 5t + 100 billion dollars, where t is years after 2000.
N'(t) = 2t + 5
In 2010 (t = 10): N'(10) = 2(10) + 5 = 25 billion dollars per year.
The GDP was 25 billion dollars per year at that point.
Percentage rate of change: 100 · N'(t)/N(t) = 100 · (25/250) = approximately 9.8%.
\frac{d}{dx}(c) = 0 \quad \text{(constant rule)}\frac{d}{dx}(x^n) = n \cdot x^{n-1} \quad \text{(power rule, any real } n\text{)}\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x) \quad \text{(constant multiple rule)}\frac{1}{x^n} = x^{-n} \quad \text{(algebraic rewrite for negative exponents)}\frac{a \cdot b}{c} = a \cdot \frac{b}{c} \quad \text{(pulling constants out of fractions)}Students often think the derivative of a constant is 1. It is zero. The derivative of x^1 is 1 (by the power rule), which is a different thing entirely.
Students sometimes forget to rewrite terms before differentiating. You cannot apply the power rule directly to 1/x^3 without first rewriting it as x^(-3).
When using the constant multiple rule, students occasionally differentiate the constant as well as the function. The constant stays fixed; only the function part gets differentiated.
Students confuse the power rule with multiplication: the derivative of x^4 is 4x^3, not 4 times x^4. The exponent drops by one.
These three rules appear in nearly every derivative problem on the exam. If you cannot apply them quickly and accurately, every later topic (product rule, quotient rule, chain rule) will compound the difficulty.
Percentage rate of change questions (like the GDP example) are a favourite exam format. The formula is 100 · f'(x)/f(x). Know it cold.
Expect exam questions that give you messy expressions with roots and fractions in the denominator. The test is whether you can rewrite them into power-rule form before differentiating.
True or false: The derivative of 7 is 7.
Answer: False. The derivative of any constant is 0.
Fill in the blank: The derivative of x^5 is ___.
Answer: 5x^4.
True or false: The derivative of 3x^2 equals 3 times the derivative of x^2.
Answer: True. That is exactly the constant multiple rule.
Fill in the blank: To differentiate 1/x^6, first rewrite it as ___.
Answer: x^(-6).
True or false: The percentage rate of change of f(x) at a point is f'(x)/f(x).
Answer: False. It is 100 · f'(x)/f(x). Do not forget the factor of 100.
Q: Find f'(x) for f(x) = 4x^7 - 3x^2 + 9.
A: f'(x) = 28x^6 - 6x. The constant 9 drops out.
Q: Find f'(x) for f(x) = (2/x^3) + 5x.
A: Rewrite as 2x^(-3) + 5x. Then f'(x) = 2(-3)x^(-4) + 5 = -6x^(-4) + 5 = -6/x^4 + 5.
Q: The revenue of a company is R(t) = t^2 + 8t + 200 million dollars, t years after 2015. What is the rate of change of revenue in 2020, and what is the percentage rate of change?
A: R'(t) = 2t + 8. In 2020, t = 5, so R'(5) = 18 million dollars per year. R(5) = 25 + 40 + 200 = 265. Percentage rate = 100 · (18/265) = approximately 6.8%.
Q: Differentiate f(x) = (1/4)x^(2/3) - x^(-1/2).
A: f'(x) = (1/4)(2/3)x^(-1/3) - (-1/2)x^(-3/2) = (1/6)x^(-1/3) + (1/2)x^(-3/2).
These rules feed directly into the product rule (Chapter 2, companion guide), where you will differentiate products of two functions, each of which you handle with the power rule and constant multiple rule.
The chain rule (coming later) extends the power rule to composite functions. If you are comfortable rewriting expressions and applying the power rule here, the chain rule will feel like one extra step on top.
Percentage rate of change connects to marginal analysis in business calculus, where you interpret the derivative as marginal cost, marginal revenue, or marginal profit.
Derivative, differentiation, constant rule, power rule, constant multiple rule, negative exponent, fractional exponent, polynomial derivative, rate of change, percentage rate of change, marginal analysis, d/dx, f'(x), calculus for business, FIU calculus, Chapter 2 derivatives, basic differentiation, algebraic rewrite, exponent rules