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IB DP Maths AA SL Study Notes

5.1.2 Product and Quotient Rules

Product Rule

When you're dealing with a function that's the product of two other functions, the Product Rule is your go-to method for differentiation.

Formula

If you're working with two functions, u(x) and v(x), and you want to differentiate their product, u(x) * v(x), the formula for the Product Rule is:

Derivative = derivative of u(x) times v(x) + u(x) times derivative of v(x).

This formula essentially states that to differentiate a product of two functions, you take the derivative of the first function and multiply it by the second function, then add the first function multiplied by the derivative of the second function. Understanding the domain and range of these functions is crucial for accurate differentiation.

Applications

The Product Rule is a versatile tool, applicable in a plethora of scenarios:

  • Polynomial Functions: When dealing with polynomial functions that are products of binomials or trinomials, the Product Rule simplifies the differentiation process. You might also need to consider the basics of rational functions when working with more complex polynomials.
  • Physics: In physics, especially in mechanics, quantities are often products of two functions, like force and displacement. The Product Rule can be used to differentiate such functions to determine rates of change. This can also involve the Power Rule for simpler expressions.
  • Economics: In economics, the Product Rule can be used to differentiate functions representing product-related scenarios, like cost and production levels.

Example 1:

Consider a function representing the area of a rectangle, where one side is 2x + 3 and the other is x2 - 4. The area function is f(x) = (2x + 3)(x2 - 4). To find the rate at which the area changes with respect to x, we differentiate using the Product Rule.

Letting u(x) = 2x + 3 and v(x) = x2 - 4, the derivative is:

f'(x) = 2 times (x2 - 4) + (2x + 3) times 2x.

This gives:

f'(x) = 2x2 - 8 + 4x2 + 6x.

Combining like terms:

f'(x) = 6x2 + 6x - 8.

Quotient Rule

For functions that are quotients of two other functions, the Quotient Rule is the differentiation technique of choice.

Formula

For two functions, u(x) and v(x), the derivative of their quotient, u(x) / v(x), where v(x) is not equal to zero, is:

Derivative = (derivative of u(x) times v(x) - u(x) times derivative of v(x)) divided by v(x) squared.

This formula can be thought of as the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. For better results, one should also be familiar with techniques for solving quadratic equations.

Applications

The Quotient Rule is essential in various mathematical and real-world contexts:

  • Rational Functions: For functions that are ratios of two polynomials, the Quotient Rule offers a direct method for differentiation.
  • Engineering: In engineering problems, especially in fluid dynamics and thermodynamics, ratios of functions often arise. The Quotient Rule aids in finding rates of change in such scenarios.
  • Biology: In biology, especially in growth models and enzyme kinetics, the Quotient Rule can be used to differentiate functions representing ratios of quantities. Complex derivatives often require combining the Quotient Rule with other rules like the Chain Rule.

Example 2:

Consider a function representing the ratio of two quantities, say the concentration of a substance in a solution over time. The function is g(x) = (3x2 + 4) / (x - 2). To determine the rate at which the concentration changes with respect to time, we differentiate using the Quotient Rule.

Letting u(x) = 3x2 + 4 and v(x) = x - 2, the derivative is:

g'(x) = (6x times (x - 2) - (3x2 + 4) times 1) divided by (x - 2) squared.

Simplifying:

g'(x) = (6x2 - 12x - 3x2 - 4) divided by (x2 - 4x + 4).

Combining terms:

g'(x) = (3x2 - 12x - 4) divided by (x2 - 4x + 4).

FAQ

Yes, the Product Rule can be extended to differentiate the product of more than two functions. If you have a product of three functions, for instance, you'd differentiate one function at a time while keeping the other two constant, and then sum up the results. This process can be extended to any number of functions. However, the more functions involved, the more terms you'll have in the result, making the process more complex.

A common mistake is mixing up the terms in the Product and Quotient Rules. For the Product Rule, students sometimes forget to add the two terms together. For the Quotient Rule, the order of the terms in the numerator is crucial; subtracting in the wrong order can lead to incorrect results. Another common error is forgetting to square the denominator in the Quotient Rule. Regular practice and careful attention to the formulas can help avoid these pitfalls.

The Product and Quotient Rules are just two of the many differentiation techniques in calculus. They are foundational and often introduced after students learn basic differentiation concepts. As students progress, they'll encounter other rules and techniques, like the Chain Rule, Implicit Differentiation, and techniques for integration. Understanding the Product and Quotient Rules is crucial as they form the basis for more advanced topics and have wide-ranging applications in various fields, from physics to economics.

The Product and Quotient Rules are indispensable because not all functions we encounter are simple or can be easily separated. In real-world scenarios, functions often come in the form of products or quotients of two or more simpler functions. While we could expand or simplify these functions in some cases, doing so can be tedious or even impossible in others. The Product and Quotient Rules provide a systematic approach to differentiate such functions directly without the need for simplification, making the process more efficient and manageable.

The Quotient Rule can be derived from the Product Rule and the Chain Rule. If you have a function in the form u(x) / v(x), you can rewrite it as u(x) * v(x)-1. Differentiating this using the Product Rule and the Chain Rule will give you the Quotient Rule. This connection showcases the consistency within differentiation rules and how they can be interrelated.

Practice Questions

ifferentiate the function f(x) = (x^3 + 5x^2) / (2x - 3).

To differentiate the function, we'll use the quotient rule. The quotient rule states that if you have a function in the form u(x) / v(x), then its derivative is given by:

f'(x) = (u'v - uv') / v2

Here, u(x) = x3 + 5x2 and v(x) = 2x - 3.

Differentiating u with respect to x, we get u' = 3x2 + 10x.

Differentiating v with respect to x, we get v' = 2.

Plugging these values into the quotient rule formula, we get:

f'(x) = (3x2 + 10x)(2x - 3) - (x3 + 5x2)(2) / (2x - 3)2

Simplifying this expression, we get:

f'(x) = x(4x2 + x - 30) / (2x - 3)2

Differentiate the function g(x) = (3x^2 - 4x)(x^2 + 2).

To differentiate this function, we'll use the product rule. The product rule states that if you have a function in the form u(x) * v(x), then its derivative is given by:

g'(x) = u'v + uv'

Here, u(x) = 3x2 - 4x and v(x) = x2 + 2.

Differentiating u with respect to x, we get u' = 6x - 4.

Differentiating v with respect to x, we get v' = 2x.

Plugging these values into the product rule formula, we get:

g'(x) = (6x - 4)(x2 + 2) + (3x2 - 4x)(2x)

Simplifying this expression, we get:

g'(x) = 4(3x3 - 3x2 + 3x - 2)

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
LinkedIn
Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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