Product Rule Examples: Mastering Differentiation by Parts

In the study of calculus, the product rule is a fundamental tool for differentiating functions that are formed by multiplying two or more constituent functions. It often appears in school exams, university courses, and professional applications where models involve rates, growth, or changing quantities. This article dives into product rule examples across a spectrum of complexity, from straightforward polynomials to functions that blend exponentials, trigonometric functions, and logarithms. Whether you are revising for an assessment or sharpening your intuition, the following product rule examples will help you recognise patterns, apply the rule correctly, and check your work with confidence.

What is the Product Rule?

The product rule states that the derivative of the product of two differentiable functions u(x) and v(x) is given by (uv)’ = u’v + uv’. In words: the rate of change of a product comes from the sum of two contributions — the rate of change of the first factor times the second, plus the first factor times the rate of change of the second. This extends naturally to products of three or more factors, where each term differentiates one factor at a time while keeping the others intact. In British maths classrooms, you will often see the rule introduced with a few concrete examples before moving on to more challenging cases.

Quick recap: u and v

When faced with a product uv, identify the two functions u(x) and v(x). Then compute their derivatives u'(x) and v'(x). Substitute into the formula (uv)’ = u’v + uv’. For three or more factors, the rule generalises to (uvw…)’ = u’vw… + uv’W… + uvw’… and so on, with each term differentiating a single factor in turn.

Product Rule Examples: Simple Polynomial Case

Example 1: Differentiating (3x^2)(x + 1)

Let u(x) = 3x^2 and v(x) = x + 1. Then u'(x) = 6x and v'(x) = 1. Applying the product rule, (uv)’ = u’v + uv’ gives:

• u’v = (6x)(x + 1) = 6x^2 + 6x
• uv’ = (3x^2)(1) = 3x^2

Summing the two contributions yields:

dy/dx = (uv)’ = 6x^2 + 6x + 3x^2 = 9x^2 + 6x

Factorising slightly for neatness, dy/dx = 3x(3x + 2). This is a classic product rule example that demonstrates how the derivative spreads across both factors rather than simply differentiating one and keeping the other fixed.

Example 2: Differentiating (2x + 5)(x^2)

Set u(x) = 2x + 5 and v(x) = x^2. Then u'(x) = 2 and v'(x) = 2x. Using the product rule, (uv)’ = u’v + uv’:

• u’v = 2(x^2) = 2x^2
• uv’ = (2x + 5)(2x) = 4x^2 + 10x

Thus dy/dx = 2x^2 + 4x^2 + 10x = 6x^2 + 10x. Factoring, dy/dx = 2x(3x + 5). A straightforward instance where both factors contribute to the slope in a clear and linear fashion.

Product Rule Examples: Exponential and Polynomial

Example 3: Differentiating x e^{2x}

Let u(x) = x and v(x) = e^{2x}. Then u'(x) = 1 and v'(x) = 2e^{2x} by the chain rule. The product rule gives:

• u’v = 1 · e^{2x} = e^{2x}
• uv’ = x · 2e^{2x} = 2xe^{2x}

Therefore dy/dx = e^{2x} + 2xe^{2x} = e^{2x}(1 + 2x). This example shows how the exponential function’s inner derivative via the chain rule blends with the outer product structure.

Example 4: Differentiating (x^3)(e^{x})

Take u(x) = x^3 and v(x) = e^{x}. Then u'(x) = 3x^2 and v'(x) = e^{x}. The product rule yields:

• u’v = 3x^2 e^{x}
• uv’ = x^3 e^{x}

Hence dy/dx = 3x^2 e^{x} + x^3 e^{x} = e^{x}(3x^2 + x^3) = x^2 e^{x}(3 + x). This example highlights how the derivative can factor nicely to reveal hidden structure.

Product Rule Examples: Trigonometric and Polynomial

Example 5: Differentiating x cos x

Let u(x) = x and v(x) = cos x. Then u'(x) = 1 and v'(x) = -sin x. The product rule gives:

• u’v = 1 · cos x = cos x
• uv’ = x · (-sin x) = -x sin x

Thus dy/dx = cos x – x sin x. This is a quintessential product rule example where a polynomial multiplies a trigonometric function, producing a derivative that mixes both components in a non-trivial way.

Example 6: Differentiating (sin x)(x^2)

With u(x) = sin x and v(x) = x^2, we have u'(x) = cos x and v'(x) = 2x. The derivative is:

• u’v = cos x · x^2 = x^2 cos x
• uv’ = sin x · 2x = 2x sin x

So dy/dx = x^2 cos x + 2x sin x. This example shows how the rates of change of both factors contribute with differing trigonometric and polynomial weighting.

Product Rule and the Chain Rule: Interplay in Real Problems

Example 7: Differentiating x e^{x^2}

Here u(x) = x and v(x) = e^{x^2}. Then u'(x) = 1 and v'(x) = e^{x^2} · 2x (by the chain rule). The product rule gives:

• u’v = 1 · e^{x^2} = e^{x^2}
• uv’ = x · (e^{x^2} · 2x) = 2x^2 e^{x^2}

Hence dy/dx = e^{x^2} + 2x^2 e^{x^2} = e^{x^2}(1 + 2x^2). The example emphasises how chain rule factors propagate through the product rule, creating terms that reflect both the inner and outer derivatives.

Example 8: Differentiating x(ln(1+x^2))

Take u(x) = x and v(x) = ln(1 + x^2). Then u'(x) = 1 and v'(x) = (1/(1 + x^2)) · (2x) by the chain rule. The derivative is:

• u’v = 1 · ln(1 + x^2) = ln(1 + x^2)
• uv’ = x · [2x/(1 + x^2)] = 2x^2/(1 + x^2)

Thus dy/dx = ln(1 + x^2) + 2x^2/(1 + x^2). This example demonstrates how the product rule interacts with a logarithmic function inside the second factor, producing a rational expression following simplification.

Product Rule with Three or More Factors

When a function involves three or more multiplicative factors, the product rule extends to a sum of derivatives, each differentiating a single factor while leaving the others unchanged. A common three-factor example is f(x) = x^2 e^x cos x. Differentiating requires applying the rule to each factor in turn and summing the results.

Example 9: Differentiating f(x) = x^2 e^x cos x

Let u = x^2, v = e^x, w = cos x. Then:

• u’vw = (2x)(e^x)(cos x) = 2x e^x cos x
• uv’w = (x^2)(e^x)(cos x) = x^2 e^x cos x
• uvw’ = (x^2)(e^x)(-sin x) = -x^2 e^x sin x

Summing the three gives:

dy/dx = f'(x) = 2x e^x cos x + x^2 e^x cos x – x^2 e^x sin x

Factoring e^x where possible yields:

dy/dx = e^x [2x cos x + x^2 cos x – x^2 sin x]. This three-factor example illustrates how a product with multiple layers accumulates several derivative terms, each reflecting a different pathway of change within the product.

Common Mistakes and How to Avoid Them

Mistake 1: Differentiating only one factor

One frequent error is to differentiate just the first factor (or the second) and leave the other intact, effectively treating the product as if only one part changes. Remember: the product rule accounts for the variation of both, and neglecting this leads to incorrect results.

Mistake 2: Forgetting the derivative of the second factor

Another common pitfall is to forget to differentiate the second factor or to miscalculate its derivative, especially when v’ involves the chain rule. Always check that v’ is correct and that you have included uv’ properly in the final sum.

Mistake 3: Misplacing plus signs

The term u’v and the term uv’ must be added; a sign or placement error can invert the answer. A good habit is to write a quick table listing the differentiated terms and then summing them explicitly.

Practical Tips for Memorising the Product Rule

Tip 1: Visualise the rule as two contributions

Think of the product rule as extracting the rate of change from each factor. The derivative is the sum of the rate of change of the first times the second, plus the first times the rate of change of the second. This mental picture helps when you tackle more elaborate products.

Tip 2: Practice with diverse function types

Use a range of u and v—polynomials, exponentials, trigonometric, and logarithmic functions. The more variety you encounter, the quicker you recognise patterns and the easier it becomes to identify u and v in unfamiliar problems.

Tip 3: Check your result via the product rule in reverse

For a given derivative candidate, try to factor out common elements or to reconstruct the expression as a product of two functions whose derivatives you recognise. This reverse check often catches sign errors or missing terms.

Practice Problems and Solutions

Problem 1: Differentiate (4x^2 + 3)(e^x)

Choose u = 4x^2 + 3 and v = e^x. Then u’ = 8x and v’ = e^x. Compute:

• u’v = 8x e^x
• uv’ = (4x^2 + 3) e^x

Sum: dy/dx = e^x(8x) + e^x(4x^2 + 3) = e^x(4x^2 + 8x + 3).

Problem 2: Differentiate (x^2 + 1)(x – 4)

Let u = x^2 + 1, v = x – 4. Then u’ = 2x, v’ = 1. Apply the rule:

• u’v = 2x(x – 4) = 2x^2 – 8x
• uv’ = (x^2 + 1)(1) = x^2 + 1

dy/dx = (2x^2 – 8x) + (x^2 + 1) = 3x^2 – 8x + 1.

Problem 3: Differentiate f(x) = (x^3)(cos x)

Set u = x^3 and v = cos x. Then u’ = 3x^2 and v’ = -sin x. The derivative is:

• u’v = 3x^2 cos x
• uv’ = x^3(-sin x) = -x^3 sin x

dy/dx = 3x^2 cos x – x^3 sin x.

Problem 4: Differentiate f(x) = (x^2 e^{x})(\sin x)

Here we have a three-factor product: u = x^2, v = e^{x}, w = sin x. Then:

• u’vw = (2x)(e^{x})(sin x) = 2x e^{x} sin x
• uv’w = (x^2)(e^{x})(sin x) = x^2 e^{x} sin x
• uvw’ = (x^2)(e^{x})(cos x) = x^2 e^{x} cos x

Sum: dy/dx = e^{x}[2x sin x + x^2 sin x + x^2 cos x]. This problem highlights how to manage multiple factors by treating each factor in turn.

Why the Product Rule Matters in Higher Mathematics

The product rule is not merely a calculator trick; it underpins a wide range of mathematical modelling across physics, engineering, economics, and computer science. In physics, for instance, the rate of work done by a force F(x) along a path often takes the form of a product of position-dependent terms. In electrical engineering, signals can be represented as products of time-varying envelopes and carrier waves, necessitating the product rule for accurate differentiation. In economics, marginal analysis may involve differentiating products of price and quantity functions. Mastery of product rule examples builds a robust toolkit that translates across disciplines.

Three or More Factors: A Practical Mental Model

When facing a function with three factors, remember the general expansion:

If f(x) = u(x) v(x) w(x), then f'(x) = u'(x) v(x) w(x) + u(x) v'(x) w(x) + u(x) v(x) w'(x). This approach scales to any number of factors, although the arithmetic can become lengthy. In such cases, it helps to write down the derivative in pieces and combine them at the end, ensuring no term is omitted.

A Quick Reference: Patterns in Product Rule Examples

To speed up problem solving, keep a mental catalogue of common patterns you have seen in product rule examples:

• When one factor is a polynomial and the other is an exponential function, expect terms that feature both e^{ax} and powers of x.
• Products involving trigonometric functions often produce a mixture of sine and cosine terms, sometimes multiplied by polynomials.
• Logarithmic factors introduce quotients of derivatives; remember the chain rule inside the logarithm’s derivative.
• Three-factor products yield the sum of three terms, each differentiating a different factor.

Final Thoughts: Product Rule Examples and Mastery

Product rule examples form a cornerstone of calculus mastery. By practising a range of cases—from simple polynomials to combinations of exponentials, trigonometric functions, and logarithms—you build a flexible intuition for differentiation that stands up to complex real-world problems. As you work through more product rule examples, you will notice identical structure across different contexts: identify u and v, compute u’ and v’, apply (uv)’ = u’v + uv’, and then simplify carefully. This structured approach is the quickest route to accuracy and confidence in your calculus journey.

Glossary of Key Ideas for Efficient Practice

To help you remember, here is a concise glossary tied to the product rule:

• Product rule: (uv)’ = u’v + uv’
• Generalisation: For three factors, (uvw)’ = u’vw + uv’w + uvw’
• Chain rule interaction: When inner functions are involved, differentiate inside the inner function and multiply by the derivative of the inner function as needed
• Three-factor example: A,b,c differentiated in turn yields three contributing terms

Whether you are revising for an exam or building deeper mathematical fluency, these product rule examples offer a thorough road map. Practice consistently, verify your results, and you will develop both speed and accuracy in differentiating products of functions. product rule examples