✖️ Product Rule Calculator

Calculate derivatives of products of functions with step-by-step solutions using (f·g)' = f'g + fg'

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What Is the Product Rule? A Complete Calculus Guide

The product rule is one of the most fundamental differentiation formulas in calculus. It answers the question: how do you find the derivative of a function that is the product of two other functions? Whether you are differentiating x²·sin(x), eˣ·cos(x), or x·ln(x), the product rule is the key.

Formally, if h(x) = f(x)·g(x), the product rule formula states:

The Product Rule Formula
h'(x) = f'(x)·g(x) + f(x)·g'(x)

In plain language: differentiate the first function and multiply by the second unchanged, then add the first function unchanged multiplied by the derivative of the second. A useful memory device taught in classrooms worldwide is the phrase "derivative of first times second plus first times derivative of second" — or, in the notation of the UV rule sometimes used in British and South Asian curricula: (UV)' = U'V + UV'.

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Want to go deeper? Read our full Product Rule Explained guide for an intuitive walkthrough, or explore our comprehensive Product Rule Complete Guide covering every variation and edge case.

The product rule is not an arbitrary formula — it is a direct consequence of the limit definition of the derivative. It was first published by Gottfried Wilhelm Leibniz in 1684, who noticed that the differential of a product dh = d(fg) is not simply df·dg but rather g·df + f·dg. Isaac Newton arrived at the same result independently through his fluxion calculus. This rule laid a cornerstone of differential calculus that continues to be applied daily in physics, engineering, and data science.

Why You Cannot Simply Multiply the Derivatives

The single most common misconception in introductory calculus is writing (fg)' = f'·g'. This is always wrong. A simple counterexample proves it: let h(x) = x·x = x². The correct derivative is h'(x) = 2x. Multiplying the individual derivatives gives 1·1 = 1, which is completely incorrect. The product rule correctly gives 1·x + x·1 = 2x, matching the power rule.

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Critical Error to Avoid: (f·g)' ≠ f'·g'. The correct product rule formula always produces two terms: (f·g)' = f'·g + f·g'. This is the most frequently tested conceptual error in calculus exams worldwide.

The Generalised Product Rule for Three or More Functions

The product rule extends naturally to products of three or more functions. For h(x) = f(x)·g(x)·k(x):

Generalised Product Rule — Three Functions
h'(x) = f'·g·k + f·g'·k + f·g·k'

The pattern is elegant: differentiate one factor at a time while leaving all other factors unchanged, then sum all resulting terms. The number of terms in your derivative equals the number of factors in the original product. This rule is sometimes called the Leibniz product rule and extends further to the Leibniz rule for higher derivatives, which is the generalised nth-derivative formula for products used in differential equations and quantum mechanics.

Our product rule calculator automatically detects whether your expression has two, three, or more multiplicative factors by traversing the abstract syntax tree of the parsed expression. It then displays the appropriate generalised formula in every step of the solution.

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Professor's Insight: Before calculating, count your factors. That count tells you exactly how many terms your final derivative will contain — a quick self-check that catches most errors before they happen.

Product Rule Examples — Five Worked Problems

Mastering the product rule requires practising with varied function types. The five examples below progress from basic polynomial-trigonometric products to triple-factor expressions requiring both the product rule and the chain rule simultaneously.

Example 1 — Polynomial × Trigonometric
Find the derivative of h(x) = x² · sin(x)
Identify factors: f(x) = x², g(x) = sin(x)
Differentiate each: f'(x) = 2x    g'(x) = cos(x)
Apply product rule: h'(x) = 2x·sin(x) + x²·cos(x)
Final answer: h'(x) = 2x·sin(x) + x²·cos(x)
Example 2 — Exponential × Trigonometric
Find the derivative of h(x) = eˣ · cos(x)
Identify factors: f(x) = eˣ, g(x) = cos(x)
Differentiate each: f'(x) = eˣ    g'(x) = −sin(x)
Apply product rule: h'(x) = eˣ·cos(x) + eˣ·(−sin(x))
Simplify (factor eˣ): h'(x) = eˣ(cos(x) − sin(x))
Example 3 — Polynomial × Natural Logarithm
Find the derivative of h(x) = x · ln(x)
Identify factors: f(x) = x, g(x) = ln(x)
Differentiate each: f'(x) = 1    g'(x) = 1/x
Apply product rule: h'(x) = 1·ln(x) + x·(1/x)
Simplify: h'(x) = ln(x) + 1
Example 4 — Product Rule + Chain Rule Combined
Find the derivative of h(x) = x² · e^(2x)
Identify factors: f(x) = x², g(x) = e^(2x)
Differentiate each: f'(x) = 2x    g'(x) = 2e^(2x) (chain rule applied to g)
Apply product rule: h'(x) = 2x·e^(2x) + x²·2e^(2x)
Simplify (factor 2xe^(2x)): h'(x) = 2xe^(2x)(1 + x)
Example 5 — Three-Factor Generalised Product Rule
Find the derivative of h(x) = x³ · eˣ · sin(x)
Identify three factors: f = x³, g = eˣ, k = sin(x)
Differentiate each: f' = 3x²   g' = eˣ   k' = cos(x)
Apply generalised rule: h'(x) = f'gk + fg'k + fgk'
Substitute: h'(x) = 3x²eˣsin(x) + x³eˣsin(x) + x³eˣcos(x)
Factor x²eˣ: h'(x) = x²eˣ[3sin(x) + xsin(x) + xcos(x)]
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Practice Strategy: Work each example by hand first, then enter it into the calculator above and compare your steps against the generated solution. Discrepancies will almost always reveal a sign error, a forgotten chain rule application, or an incomplete simplification.

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Ready to Practice?

Download our free Product Rule Worksheet — 75+ graded problems with full answer keys, covering all function types from this page.

Product Rule vs Chain Rule — Know the Difference

Students frequently apply the wrong rule because they misidentify the structure of an expression. The distinction is architectural: are the functions multiplied side by side, or is one function nested inside another?

📦 Product Rule — Side by Side

Use when: h(x) = f(x) · g(x)
Functions sit next to each other as separate factors.

Formula: h'(x) = f'g + fg'

🔗 Chain Rule — Nested Inside

Use when: h(x) = f(g(x))
One function is the argument of another.

Formula: h'(x) = f'(g(x)) · g'(x)

ExpressionRule(s) RequiredReason
x² · sin(x)Product RuleTwo separate factors multiplied
sin(x²)Chain Rulex² is nested inside sin()
eˣ · cos(x)Product RuleTwo separate factors multiplied
e^(cos x)Chain Rulecos(x) is nested inside the exponent
x² · sin(x²)BothProduct rule for ×; chain rule for sin(x²)
√x · ln(x)Product RuleTwo separate factors (sqrt is a power)
x · e^(x²)BothProduct rule for ×; chain rule for e^(x²)

Combining Both Rules in One Calculation

Many intermediate calculus problems require applying both rules in the same calculation. The process is always the same: apply the product rule first to identify the terms, then apply the chain rule when differentiating any composite factor within those terms. Work from the outside inward, one layer at a time. Our calculator handles these combined applications automatically using math.js's symbolic differentiation engine, which recursively applies all necessary rules.

5 Common Product Rule Mistakes — and How to Fix Them

In over two decades of teaching calculus to university students, these are the five errors that appear most consistently on homework and exams. Recognising them in your own work is the fastest path to better marks.

Mistake 1 — Multiplying the Derivatives Instead of Applying the Formula

Writing (fg)' = f'·g' is by far the most frequent error. It may arise from confusing differentiation with the power rule for products in algebra, where (fg)ⁿ = fⁿgⁿ. Differentiation does not work this way. Always write out both terms: f'g + fg'.

Mistake 2 — Omitting the Chain Rule on Composite Factors

When a factor is itself a composite function — such as sin(x²), e^(3x), or (x+1)⁴ — you must apply the chain rule when differentiating it. Forgetting the chain rule multiplier (the derivative of the inner function) is the second most common product rule error.

Mistake 3 — Applying the Product Rule When It Is Not Needed

If one of your factors is a constant, the product rule reduces to simple constant multiplication: d/dx[3·f(x)] = 3·f'(x). Students sometimes write 3·f'(x)·f(x) + 3·f(x)·1, which is incorrect. Treat constants as constants, not as functions of x.

Mistake 4 — Stopping Before Simplifying

After applying the product rule you will have two (or more) terms. Leaving the answer in the unsimplified form loses marks in most calculus courses. Always check whether you can factor out a common expression. For example, 2xeˣ + x²eˣ should be written as xeˣ(x+2).

Mistake 5 — Sign Errors with Trigonometric Derivatives

Remember: the derivative of cos(x) is −sin(x) (negative). A missing negative sign when differentiating cosine is the most common source of incorrect final answers in product rule problems involving trigonometric functions.

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Exam Strategy: After computing your derivative, substitute x = 1 into both your hand-calculated answer and the calculator's result. If the numerical values match, your derivative is correct. If they differ, you have made an algebraic error somewhere in the simplification.

How to Use This Calculator & Real-World Applications

Using the Product Rule Calculator

Enter any product expression using standard notation: * for multiplication, ^ for exponents, log(x) for natural logarithm, and sqrt(x) for square root. Select your differentiation variable, then press Enter or click the button. The calculator uses the math.js symbolic engine to compute the exact derivative and renders every formula using KaTeX for crisp mathematical typesetting.

Once your result appears, three export options are available. The Print / PDF button opens your browser's print dialog — select "Save as PDF" as the destination to get a clean, advertisement-free PDF of your solution. The Save as Text button downloads a fully formatted plain-text file containing your expression, the derivative, and all five solution steps. The Copy Result button copies the expression and derivative in LaTeX notation for pasting into documents or messaging apps.

Applications of the Product Rule

Physics — Variable Mass Systems: Momentum is p(t) = m(t)·v(t). When mass varies with time (as in a rocket expelling propellant), the rate of change of momentum requires the product rule: p'(t) = m'(t)·v(t) + m(t)·v'(t). This is the foundation of the variable-mass equation of motion in Newtonian mechanics.

Economics — Marginal Revenue: Revenue is often R(x) = p(x)·q(x), the product of price per unit and quantity sold. The marginal revenue — the rate at which revenue changes as output increases — is found by differentiating this product, and it is central to every profit-maximisation problem in microeconomics.

Biology — Epidemic Models: In SIR epidemic models, the rate of new infections is proportional to β·S(t)·I(t), the product of the susceptible and infected population counts. Analysing equilibria and rates of change requires differentiating these product terms.

Engineering — Signal Processing: Amplitude-modulated signals take the form s(t) = A(t)·cos(ωt), the product of a slowly varying envelope and a high-frequency carrier. Differentiating this signal to analyse instantaneous frequency or build control systems requires the product rule.

The Leibniz Rule: The product rule extends to nth-order derivatives through the Leibniz rule: the nth derivative of a product f·g is the binomial expansion of (f + g)ⁿ where exponentiation means repeated differentiation. This formula is used in advanced differential equations, Fourier analysis, and quantum field theory.

Proof of the Product Rule — From First Principles

Understanding why the product rule works deepens intuition far beyond memorising a formula. The proof follows directly from the limit definition of the derivative and requires only one clever algebraic step — adding and subtracting the same term — that transforms an impossible limit into two manageable ones.

Let h(x) = f(x)·g(x). By definition, the derivative is:

Limit Definition Applied to a Product
h'(x) = limΔx→0 [f(x+Δx)·g(x+Δx) − f(x)·g(x)] / Δx

The key trick is to add and subtract f(x)·g(x+Δx) in the numerator:

Add and Subtract the Bridge Term
= lim [f(x+Δx)·g(x+Δx) − f(x)·g(x+Δx) + f(x)·g(x+Δx) − f(x)·g(x)] / Δx

Factor by grouping the first two and last two terms:

Factor Each Pair
= lim [g(x+Δx)·(f(x+Δx)−f(x)) + f(x)·(g(x+Δx)−g(x))] / Δx

Split over the sum and apply the limit separately to each term:

Apply the Limit — Final Result
= g(x)·f'(x) + f(x)·g'(x)

The first limit gives g(x)·f'(x) because as Δx→0, g(x+Δx)→g(x) (continuity of differentiable functions) and [f(x+Δx)−f(x)]/Δx→f'(x) (definition of derivative). The second limit gives f(x)·g'(x) for the same reason. Combined, this yields the product rule: h'(x) = f'(x)·g(x) + f(x)·g'(x).

This proof highlights that the product rule is not an approximation or a shortcut — it is an exact consequence of the limit definition, valid for all functions that are differentiable at the point in question. The same algebraic strategy (adding and subtracting a bridge term) appears in the proofs of the quotient rule and the chain rule, making it a foundational technique in real analysis.

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Professor's Note: In a first-year calculus exam, you may be asked to prove the product rule from first principles. Memorise the "add and subtract f(x)·g(x+Δx)" step — it is the single non-obvious move the entire proof depends on. Everything before and after it follows mechanically from the definition.

Product Rule with Specific Function Types

Different combinations of function types appear repeatedly in calculus courses. Understanding the patterns for each pairing accelerates both hand calculation and result verification.

Polynomial × Trigonometric Functions

These are the most common product rule problems in introductory calculus. The polynomial factor always decreases in degree after differentiation, while the trig factor cycles through sine and cosine. For h(x) = xⁿ·sin(x): differentiate xⁿ to get n·xⁿ⁻¹, differentiate sin(x) to get cos(x), and combine. The result is n·xⁿ⁻¹·sin(x) + xⁿ·cos(x). Notice that for large n, repeated differentiation of this product (using the product rule multiple times) produces results involving all powers from xⁿ down to x⁰, a pattern used in integration by parts.

Exponential × Trigonometric Functions

Products like eˣ·sin(x) and eˣ·cos(x) are the gateway to a beautiful technique in integration: the tabular method (or "DI method") applied to integration by parts. When differentiating these products, note that the exponential factor is self-replicating under differentiation — eˣ always stays as eˣ — so the product rule yields: (eˣ·sin x)' = eˣ·sin x + eˣ·cos x = eˣ(sin x + cos x). The same structure with a negative sign appears when differentiating eˣ·cos x, producing eˣ(cos x − sin x). These two derivatives are each other's counterparts and appear repeatedly in physics problems involving damped oscillations.

Polynomial × Logarithmic Functions

Products involving the natural logarithm frequently appear in information theory, thermodynamics, and statistics. The key fact to remember is that d/dx[ln x] = 1/x. Applying the product rule to xⁿ·ln x gives n·xⁿ⁻¹·ln x + xⁿ·(1/x) = n·xⁿ⁻¹·ln x + xⁿ⁻¹ = xⁿ⁻¹(n·ln x + 1). For the simple case x·ln x — which appears in entropy calculations — the derivative simplifies cleanly to ln x + 1. The function x·ln x − x is the antiderivative of ln x, which students often encounter when computing ∫ln x dx using integration by parts.

Radical × Logarithmic Functions

Products like √x·ln x = x^(1/2)·ln x combine a power-rule factor with a logarithm. Write the radical as x^(1/2) before differentiating. The product rule gives: (x^(1/2)·ln x)' = (1/2)x^(−1/2)·ln x + x^(1/2)·(1/x) = (ln x)/(2√x) + 1/√x = (ln x + 2)/(2√x). Factoring out 1/(2√x) is the clean final form. This kind of simplification — pulling out a common radical denominator — is a skill that distinguishes strong calculus students from average ones.

Products Involving Inverse Trigonometric Functions

Expressions such as x·arctan(x) or x²·arcsin(x) require knowledge of the derivatives of inverse trig functions: d/dx[arctan x] = 1/(1+x²), d/dx[arcsin x] = 1/√(1−x²). These appear often in probability distributions and geometric applications. For x·arctan x: (x·arctan x)' = arctan x + x·(1/(1+x²)) = arctan x + x/(1+x²). This result is the integrand encountered when computing ∫arctan x dx, another integration by parts pairing.

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Quick Reference — Key Derivatives to Memorise: d/dx[sin x] = cos x  ·  d/dx[cos x] = −sin x  ·  d/dx[eˣ] = eˣ  ·  d/dx[ln x] = 1/x  ·  d/dx[xⁿ] = n·xⁿ⁻¹  ·  d/dx[arctan x] = 1/(1+x²). Having these six at your fingertips makes the product rule mechanical rather than laborious.

The Product Rule in Advanced Mathematics

The product rule is not confined to first-year calculus. It appears at the heart of several advanced mathematical structures that underpin modern science and engineering.

The Leibniz Rule for Higher-Order Derivatives

The nth derivative of a product f·g is given by the Leibniz rule:

Leibniz Rule — nth Derivative of a Product
(fg)⁽ⁿ⁾ = Σk=0n C(n,k) · f⁽ᵏ⁾ · g⁽ⁿ⁻ᵏ⁾

where C(n,k) = n!/(k!(n−k)!) are the binomial coefficients. The formula is structurally identical to the binomial theorem — and this is not a coincidence. Both arise from the same combinatorial reasoning: to get the nth "derivative" (or "power"), you choose in how many ways to "act" on each factor k times out of n total actions. The Leibniz rule is used to compute high-order derivatives efficiently in differential equations and in quantum mechanics (where the nth-order correction to a Hamiltonian perturbation involves exactly this structure).

The Product Rule in Multivariable Calculus

For functions of several variables, the product rule applies independently along each partial derivative direction. If h(x,y) = f(x,y)·g(x,y), then ∂h/∂x = (∂f/∂x)·g + f·(∂g/∂x) and ∂h/∂y = (∂f/∂y)·g + f·(∂g/∂y). This is used constantly in thermodynamics, fluid mechanics, and electromagnetism when differentiating products of field quantities. The gradient of a product satisfies ∇(fg) = g·∇f + f·∇g — the vector generalisation of the product rule.

The Product Rule in Differential Geometry

The product rule for covariant derivatives on a Riemannian manifold is a foundational axiom of differential geometry: ∇(f·v) = (∇f)·v + f·∇v for a scalar function f and a vector field v. It is the requirement that the covariant derivative satisfies this "Leibniz property" that distinguishes a proper connection from an arbitrary linear map. General relativity and modern gauge theory are built on covariant derivatives that satisfy this product rule at every point of spacetime.

The Product Rule in Probability — The Score Function

In maximum likelihood estimation, the log-likelihood is the natural logarithm of a product of probability density functions. The product rule (via the log-derivative trick) converts differentiation of a product into a sum of simpler derivatives: d/dθ[ln(f·g)] = f'/f + g'/g. This decomposition is used in every gradient-based learning algorithm in machine learning, including backpropagation through product gates in neural networks.

Integration by Parts — The Inverse of the Product Rule

The most direct consequence of the product rule in integral calculus is the integration by parts formula. Starting from the product rule (fg)' = f'g + fg' and integrating both sides: ∫fg' dx = fg − ∫f'g dx. This formula allows difficult integrals to be evaluated by strategically choosing which factor to differentiate and which to integrate. It is used to compute ∫x·eˣ dx, ∫x·ln x dx, ∫x·sin x dx, ∫arctan x dx, and many others. Mastering the product rule is therefore a direct prerequisite for mastering integration by parts — arguably the single most powerful integration technique in calculus.

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Connection to Higher Mathematics: The algebraic identity that defines the product rule — d(fg) = f·dg + g·df — is the defining property of a derivation in abstract algebra. Any linear map on an algebra that satisfies this Leibniz identity is called a derivation, and the study of derivations forms the bridge between calculus and modern algebra, Lie theory, and noncommutative geometry.

Product Rule Quick Reference — 12 Common Derivatives

The table below provides instant-reference derivatives for the twelve most frequently encountered product rule combinations in calculus courses worldwide. Use these to verify calculator results or as a study guide before exams.

h(x) = f(x) · g(x)h'(x) — Product Rule Applied & Simplified
x · sin(x)sin(x) + x·cos(x)
x² · sin(x)2x·sin(x) + x²·cos(x)
x · cos(x)cos(x) − x·sin(x)
x · eˣeˣ + x·eˣ = eˣ(x + 1)
x² · eˣ2x·eˣ + x²·eˣ = xeˣ(x + 2)
eˣ · sin(x)eˣ·sin(x) + eˣ·cos(x) = eˣ(sin x + cos x)
eˣ · cos(x)eˣ·cos(x) − eˣ·sin(x) = eˣ(cos x − sin x)
x · ln(x)ln(x) + 1
x² · ln(x)2x·ln(x) + x = x(2·ln x + 1)
√x · ln(x)(ln x + 2) / (2√x)
sin(x) · cos(x)cos²(x) − sin²(x) = cos(2x)
x · arctan(x)arctan(x) + x/(1 + x²)

All twelve results above can be verified using this calculator. Enter the expression, click calculate, and compare the displayed derivative against the table. If you are preparing for an exam, cover the right column and work each derivative from scratch — then uncover and check. Discrepancies will point exactly to which rule (chain, power, or trig identity) you need to review.

Tips for Faster Product Rule Calculations

Tip 1 — Write the formula first, then substitute. Always write (fg)' = f'g + fg' on your paper before substituting. Students who substitute directly, without writing the template, skip terms far more often than those who follow the two-step process. The few extra seconds this takes will save far more time in error correction.

Tip 2 — Factor your final answer. After applying the product rule, always scan for a common factor in both terms of the result. Factoring the answer is expected in every calculus course. For xeˣ + x²eˣ, the common factor xeˣ gives xeˣ(1+x) — compact, elegant, and the correct final form.

Tip 3 — Verify numerically. After computing a derivative, plug in a simple value like x = 1 or x = 0 into both your derivative and the original function's numerical derivative (Δh/Δx for a small Δx). If the numbers agree to three significant figures, your derivative is almost certainly correct. If they disagree, you have an error in your algebra, not your method.

Tip 4 — Use logarithmic differentiation for difficult products. For products of three or more functions, or when factors involve awkward powers, the technique of logarithmic differentiation (take ln of both sides, differentiate implicitly, multiply back) often produces a cleaner calculation than applying the product rule directly. Our calculator handles these automatically, but recognising when logarithmic differentiation is more efficient is a mark of advanced calculus skill.

❓ Frequently Asked Questions

The product rule states that the derivative of a product of two functions f(x) and g(x) is: (f·g)' = f'·g + f·g'. In words, differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second. It was first published by Leibniz in 1684 and is one of the five fundamental differentiation rules alongside the sum rule, chain rule, quotient rule, and power rule.

Use the product rule when two distinct functions are multiplied together as separate factors: x²·sin(x), eˣ·cos(x). Use the chain rule when one function is nested inside another: sin(x²), e^(cos x). If your expression has both (such as x²·sin(x²)), apply the product rule to the multiplication and the chain rule when differentiating the composite inner factor sin(x²).

For two functions: (fg)' = f'g + fg'. For three functions: (fgh)' = f'gh + fg'h + fgh'. In general, the derivative of a product of n functions is the sum of n terms, each formed by differentiating exactly one factor and leaving all others unchanged. This is the Leibniz generalisation of the product rule.

Use * for multiplication between factors, ^ for exponents, and parentheses for grouping. Write log(x) for the natural logarithm ln(x), sqrt(x) for square root, and use standard names for trig functions: sin, cos, tan. Examples: x^2 * sin(x), e^x * cos(x), (x+1) * (x^2 - 3), sqrt(x) * log(x), x^3 * e^x * sin(x). Press Enter or click the button to calculate instantly.

Yes. Three export options appear below your result. "Print / PDF" opens your browser's print dialog — choose "Save as PDF" as the printer to create a clean PDF with all advertising and navigation removed. "Save as Text" downloads a plain-text file with your expression, derivative, and all five solution steps. "Copy Result" copies the derivative in LaTeX notation to your clipboard for pasting into Word, LaTeX editors, or chat apps.

Yes. For three functions f, g, and h the generalised product rule is (fgh)' = f'gh + fg'h + fgh'. Our calculator detects multi-factor products by analysing the expression's syntax tree rather than searching for a multiplication sign in the text, so it correctly identifies all factors in expressions like x³·eˣ·sin(x) and applies the full generalised formula.

The most common mistake is writing (fg)' = f'·g' — multiplying the two derivatives rather than applying the full formula. The correct rule always produces two (or more) terms added together: (fg)' = f'g + fg'. A closely related error is forgetting to apply the chain rule when one of the factors is itself a composite function like sin(x²) or e^(3x).

The proof uses the limit definition of the derivative. Write h'(x) = lim[f(x+Δx)g(x+Δx) − f(x)g(x)]/Δx, then add and subtract the bridge term f(x)·g(x+Δx) in the numerator. This splits the limit into two parts: g(x+Δx)·[f(x+Δx)−f(x)]/Δx and f(x)·[g(x+Δx)−g(x)]/Δx. As Δx→0, these converge to g(x)·f'(x) and f(x)·g'(x) respectively, giving the product rule. The "add and subtract a bridge term" technique is the key non-obvious step.

The Leibniz rule extends the product rule to the nth derivative: (fg)⁽ⁿ⁾ = Σ C(n,k)·f⁽ᵏ⁾·g⁽ⁿ⁻ᵏ⁾ where the sum runs from k=0 to k=n and C(n,k) are binomial coefficients. This is the exact analogue of the binomial theorem, with repeated differentiation replacing repeated multiplication. It is used in differential equations, quantum mechanics, and generating function analysis to compute high-order derivatives of products efficiently.

Integration by parts is derived directly from the product rule. Starting from (fg)' = f'g + fg', integrate both sides: ∫(fg)' dx = ∫f'g dx + ∫fg' dx, which gives fg = ∫f'g dx + ∫fg' dx. Rearranging: ∫fg' dx = fg − ∫f'g dx. This is the integration by parts formula. Every time you use integration by parts — to evaluate ∫x·eˣ dx, ∫x·ln x dx, or ∫arctan x dx — you are applying the product rule in reverse.