Fundamentals of Business Mathematics and Statistics · Calculus - Application in Business

Differentiation Basics and Rules for CMA Foundation

Updated 10 October 2026 · Fact-checked

Differentiation finds the rate at which a function changes. The derivative dy/dx gives the slope of the curve at a point. To solve questions, identify the function type, apply the power, sum, product, quotient or chain rule, simplify, and substitute the given value of x if asked.

Understand Differentiation Basics and Rules

A function such as cost C(x) tells you the cost for x units. Often you want to know how fast cost changes when x changes a little. That rate of change is the derivative, written dy/dx or f'(x).

Geometrically, the derivative is the slope of the tangent to the curve at a point. If the derivative is positive, the function is rising there. If it is negative, the function is falling.

You do not calculate derivatives from first principles in an objective paper. You use rules. The power rule handles terms like x³. The sum rule lets you differentiate term by term. The product and quotient rules handle functions multiplied or divided. The chain rule handles a function inside another function, such as (3x + 2)⁵ or e^(x²).

In business, the derivative of cost is marginal cost and the derivative of revenue is marginal revenue. Differentiation is also the base for maxima and minima. Learn the rules cold, because each question is a short, direct application.

Key formulas to remember

Derivative of a constant: d/dx (c) = 0; Any fixed number, such as 7 or ₹500, has zero rate of change.
Power rule: d/dx (xⁿ) = n·xⁿ⁻¹; Works for any real n, including fractions and negatives. √x = x^(1/2) and 1/x = x⁻¹.
Constant multiple and sum rule: d/dx [a·f(x) ± b·g(x)] = a·f'(x) ± b·g'(x); Differentiate each term separately.
Product rule: d/dx (u·v) = u·(dv/dx) + v·(du/dx); Use when two functions of x are multiplied.
Quotient rule: d/dx (u ÷ v) = [v·(du/dx) − u·(dv/dx)] ÷ v²; For v ≠ 0. The numerator order matters: v·u' first, then minus u·v'.
Chain rule: dy/dx = (dy/du) × (du/dx); For y = f(u) where u = g(x). Differentiate outside, then multiply by the derivative of the inside.
Exponential functions: d/dx (eˣ) = eˣ; d/dx (e^(f(x))) = e^(f(x)) · f'(x); d/dx (aˣ) = aˣ · log_e a; For a > 0. eˣ is its own derivative.
Logarithmic functions: d/dx (log_e x) = 1/x; d/dx (log_e f(x)) = f'(x) ÷ f(x); For x > 0 and f(x) > 0. Here log means natural log (base e).

How to solve Differentiation Basics and Rules questions

Use this method for any differentiation question, whether you must find the derivative or its value at a point.

1Rewrite the function so every term is in a form the rules cover: roots as powers, 1/xⁿ as x⁻ⁿ.
2Decide the structure: a sum of terms, a product, a quotient, or a function of a function.
3Pick the matching rule. For a sum, differentiate term by term. For a product use u·v' + v·u'. For a quotient use (v·u' − u·v') ÷ v². For a bracket raised to a power or e or log of an expression, use the chain rule.
4Differentiate the parts carefully, writing u, v, u' and v' separately when two rules are involved.
5Combine the parts and simplify, but only as far as needed to match an option.
6If a value of x is given, substitute it only after finding the derivative.
7Check the answer against the options. Eliminate any that have the wrong power or a missing chain factor.

Quickest way: Spot the rule, then test with a number

When to use it: Use when options look similar and time is short, especially for product, quotient and chain rule questions.

For y = (ax + b)ⁿ, write the answer directly as n·a·(ax + b)ⁿ⁻¹.
For e^(ax + b) the derivative is a·e^(ax + b). For log(ax + b) it is a ÷ (ax + b).
For a product, you can first multiply it out if it is a simple polynomial, then use the power rule.
For a quotient, if the denominator is a single term like x², split it into separate powers and avoid the quotient rule.
If options differ, substitute x = 1 (or another easy value) into the derivative and into each option to eliminate wrong ones.

Common mistakes in Differentiation Basics and Rules

Writing the derivative of eˣ as x·eˣ⁻¹.
Students apply the power rule to a function where x is the exponent, not the base.
Fix: The power rule is only for xⁿ with a fixed power. For eˣ the derivative is eˣ.
Differentiating a product as u'·v'.
It feels natural to differentiate each factor and multiply.
Fix: Use u·v' + v·u'. Test with x·x: the answer must be 2x, not 1.
Reversing the numerator in the quotient rule, giving u·v' − v·u'.
The two terms look alike and the order is easy to forget.
Fix: Remember 'low d-high minus high d-low': v·u' − u·v', all over v².
Forgetting the inner derivative in the chain rule, such as d/dx (3x + 1)⁴ = 4(3x + 1)³.
Students stop after differentiating the outside.
Fix: Always multiply by the derivative of the bracket. The correct answer is 4 × 3 × (3x + 1)³ = 12(3x + 1)³.
Writing the derivative of log x as 1/x for log of an expression, such as d/dx log(x² + 1) = 1/(x² + 1).
The rule d/dx log x = 1/x is applied without the chain factor.
Fix: Use f'(x) ÷ f(x). Here the answer is 2x ÷ (x² + 1).
Saying the derivative of a constant term like 5x² + 7 includes 7.
Students forget constants vanish.
Fix: d/dx (7) = 0. Only the 5x² term gives 10x.

Worked examples

Example 1

If y = x²·eˣ, find dy/dx at x = 1. Options: (A) e (B) 2e (C) 3e (D) 4e

Show the solution

Here the function is a product. Let u = x² and v = eˣ.
Then du/dx = 2x and dv/dx = eˣ.
By the product rule, dy/dx = u·v' + v·u' = x²·eˣ + eˣ·2x = eˣ(x² + 2x).
At x = 1: dy/dx = e(1 + 2) = 3e.

Answer: (C) 3e

Example 2

Find dy/dx for y = log_e (x² + 1) at x = 1. Options: (A) 1/2 (B) 1 (C) 2 (D) 1/4

Show the solution

This is a log of an expression, so use the chain rule: dy/dx = f'(x) ÷ f(x).
Here f(x) = x² + 1, so f'(x) = 2x.
dy/dx = 2x ÷ (x² + 1).
At x = 1: dy/dx = 2 ÷ 2 = 1.

Answer: (B) 1

Exam tips

Questions are usually one-step or two-step applications. Identify the rule within five seconds and move on.
Keep the options in view. Many wrong options come from a missed chain factor or a reversed quotient rule.
Substitute the given x only after differentiating. Substituting first turns the function into a constant with derivative zero.
Convert roots and fractions to powers before differentiating, for example 1/√x = x^(−1/2).
With no negative marking, always mark an answer. If stuck, eliminate options using a simple test value.

Practice questions from Calculus - Application in Business

Differentiation Basics and Rules in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

CA FoundationDifferentiation Rules and Standard Formulas

Differentiation Basics and Rules: frequently asked questions

What is the derivative of a function in simple terms?

The derivative is the rate at which a function changes as x changes. It also equals the slope of the tangent to the curve. In business it appears as marginal cost or marginal revenue.

Which differentiation formulas must I memorise for CMA Foundation?

Learn the power, sum, product, quotient and chain rules. Also learn the derivatives of eˣ, aˣ and log x. These cover almost all questions on this topic.

How do I differentiate exponential and logarithmic functions?

For e^(f(x)), the derivative is e^(f(x)) multiplied by f'(x). For log f(x), it is f'(x) divided by f(x). The chain factor f'(x) is the part students most often miss.

When should I use the chain rule instead of the product rule?

Use the chain rule when one function sits inside another, like (2x + 3)⁵ or e^(x²). Use the product rule when two separate functions of x are multiplied, like x·eˣ.