Why We Need Rules

Computing every derivative from the limit definition is correct but brutally slow. Calculus provides a toolkit of rules — each proved once from the definition, then usable forever. Learn these four rules fluently and you can differentiate almost anything instantly.

The Big Four

Power Rule — for xⁿ terms  ·  Sum Rule — for adding functions  ·  Product Rule — for multiplying functions  ·  Quotient Rule — for dividing functions. The Chain Rule (next article) handles compositions.

Power Rule
d/dx [xⁿ] = nxⁿ⁻¹

The most-used rule in calculus. Bring the exponent down as a coefficient, then reduce the exponent by 1. Works for any real exponent n — positive, negative, fractional.

d/dx[x⁵]= 5x⁴
d/dx[x⁻²]= −2x⁻³ = −2/x³
d/dx[√x]= d/dx[x^(1/2)] = (1/2)x^(−1/2) = 1/(2√x)
d/dx[7]= 0 (constants have zero derivative — the special case n=0)
Sum & Difference Rules
(f ± g)' = f' ± g'

Differentiate term by term. The derivative of a sum is the sum of derivatives. This is what makes polynomials so easy to differentiate — just apply the Power Rule to each term.

Exampled/dx[4x³ − 7x² + 2x − 5] = 12x² − 14x + 2
Product Rule
(fg)' = f'g + fg'

For the product of two functions: first times derivative of second, plus second times derivative of first. A common mnemonic: "first d-second + second d-first."

ProblemDifferentiate h(x) = x² · sin(x)
Step 1f = x², g = sin(x). So f' = 2x, g' = cos(x).
Step 2h'(x) = f'g + fg' = 2x·sin(x) + x²·cos(x)

⚠️ Common mistake: (fg)' ≠ f'g'. You cannot just multiply the derivatives.

Quotient Rule
(f/g)' = (f'g − fg') / g²

For a ratio of two functions. Mnemonic: "lo d-hi minus hi d-lo, over lo squared" (lo = denominator g, hi = numerator f).

ProblemDifferentiate h(x) = x³ / (x² + 1)
Step 1f = x³, g = x²+1. f' = 3x², g' = 2x.
Step 2h'(x) = (3x²(x²+1) − x³·2x) / (x²+1)²
Step 3= (3x⁴ + 3x² − 2x⁴) / (x²+1)² = (x⁴ + 3x²) / (x²+1)²
Complete Derivative Cheat Sheet
d/dx [c]= 0
d/dx [xⁿ]= nxⁿ⁻¹
d/dx [eˣ]= eˣ
d/dx [aˣ]= aˣ ln(a)
d/dx [ln x]= 1/x
d/dx [sin x]= cos x
d/dx [cos x]= −sin x
d/dx [tan x]= sec²x
d/dx [arcsin x]= 1/√(1−x²)
d/dx [arctan x]= 1/(1+x²)

Why Rules, Not the Limit Definition Every Time?

The limit definition of the derivative is the rigorous foundation — but computing d/dx[x¹⁰⁰] from first principles would require expanding (x+h)¹⁰⁰ using the binomial theorem, a page-long calculation. The Power Rule gives the answer in one step: 100x⁹⁹. Each differentiation rule was proved once from the limit definition, then enshrined as a shortcut valid for all qualifying functions. Learning the rules — and their conditions — is what makes calculus computationally practical.

Proving the Product Rule

The Product Rule is not obvious, and understanding its proof deepens your ability to use it correctly. Starting from the limit definition: (fg)'(x) = lim(h→0) [f(x+h)g(x+h) − f(x)g(x)] / h. Add and subtract f(x+h)g(x) in the numerator: = lim [f(x+h)g(x+h) − f(x+h)g(x) + f(x+h)g(x) − f(x)g(x)] / h = lim f(x+h)·[g(x+h)−g(x)]/h + lim g(x)·[f(x+h)−f(x)]/h. As h→0: f(x+h) → f(x) (continuity), [g(x+h)−g(x)]/h → g'(x), [f(x+h)−f(x)]/h → f'(x). Result: f(x)g'(x) + g(x)f'(x) = f'g + fg'. This is why the rule takes the form it does.

Extending the Product Rule to Three or More Functions

For a product of three functions: (fgh)' = f'gh + fg'h + fgh'. The pattern: differentiate each factor once, leaving all others unchanged, and sum the results. For n functions: (f₁f₂···fₙ)' = Σᵢ f₁···fᵢ'···fₙ (differentiate one factor at a time, sum over all choices). This generalisation is occasionally needed for complex expressions and is easily derived by applying the two-function Product Rule iteratively.

Combining Multiple Rules — Complex Examples

Example 1 d/dx[x²·sin(x)·eˣ] — Three Functions
StrategyApply the Product Rule to (x²·sin x)·eˣ first, then apply it again inside.
Step 1Let u = x²sin x, v = eˣ. u' = 2x sin x + x²cos x (Product Rule). v' = eˣ.
Result(2x sin x + x²cos x)eˣ + x²sin x·eˣ = eˣ(2x sin x + x²cos x + x²sin x).
Example 2 d/dx[sin(x²)/(x³+1)] — Quotient + Chain
Numeratord/dx[sin(x²)] = 2x·cos(x²) — Chain Rule needed.
Denominatord/dx[x³+1] = 3x².
Quotient Rule[2x·cos(x²)·(x³+1) − sin(x²)·3x²] / (x³+1)².

The Constant Multiple and Sum Rules — Why They Work

d/dx[cf(x)] = c·f'(x) follows from the limit definition: c·f(x) changes at rate c times the rate of f(x). d/dx[f+g] = f'+g' follows because limits distribute over sums. These two rules together mean differentiation is a linear operator — a concept that becomes central in differential equations and functional analysis. In matrix form for polynomials, differentiation can be represented as a matrix acting on the vector of coefficients: d/dx[a₀ + a₁x + a₂x² + a₃x³] = a₁ + 2a₂x + 3a₃x², which corresponds to multiplying the coefficient vector by the "differentiation matrix."

When Rules Break Down

The Product, Quotient, and Power Rules all require the functions to be differentiable at the point in question. The Power Rule d/dx[xⁿ] requires n to be a constant — it does not apply to xˣ (where the exponent also depends on x). For xˣ, use logarithmic differentiation: ln y = x ln x → (1/y)(dy/dx) = ln x + 1 → dy/dx = xˣ(ln x + 1). The Quotient Rule requires g(x) ≠ 0 at the point of differentiation. Always check these conditions before applying rules mechanically.

Frequently Asked Questions
When do I use the Product Rule vs the Chain Rule?
Product Rule: when two separate functions are multiplied, e.g., x²·sin(x). Chain Rule: when one function is nested inside another, e.g., sin(x²). The key question: are the functions multiplied together (Product Rule) or is one the input to the other (Chain Rule)? Sometimes both apply simultaneously.
Can I use the Quotient Rule as Product Rule with a negative exponent?
Yes — f/g = f · g⁻¹, and you can differentiate using the Product Rule combined with the Chain Rule: (f · g⁻¹)' = f'·g⁻¹ + f·(−1)g⁻²·g'. Both methods give the same answer. Many experienced calculus users prefer this approach as it avoids memorising a separate formula.
Does the Power Rule work for negative and fractional exponents?
Yes — the Power Rule d/dx[xⁿ] = nxⁿ⁻¹ works for any real exponent n. d/dx[x⁻³] = −3x⁻⁴. d/dx[x^(2/3)] = (2/3)x^(−1/3). Always rewrite roots as fractional exponents first: √x = x^(1/2), ∛x = x^(1/3).
← Previous
What is a Derivative?
Next →
Chain Rule Explained
References & Further Reading
  • Stewart, J. (2015). Calculus, §3.1–3.4. Cengage.
  • Apostol, T. (1967). Calculus, Vol. 1, Ch. 4. Wiley.
  • Larson, R. & Edwards, B. (2013). Calculus, §2.2–2.5. Cengage.

To see all these rules tested in exam conditions, see the AP Calculus Review guide. Work through practice problems to build fluency, and consult the Common Mistakes guide to avoid the errors that cost marks.

AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus at university level

Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students at both undergraduate and postgraduate level. Her research focuses on numerical methods for differential equations. She has reviewed this article for mathematical accuracy and pedagogical clarity.

Technically reviewed by: Prof. James Chen, Stanford Mathematics Department