Product Rule With Three Functions

Unveiling the Mystery: Product Rule with Three (or More!) Functions

The product rule of differentiation is a fundamental concept in calculus. Most introductory courses cover the rule for two functions: d/dx[f(x)g(x)] = f'(x)g(x) + f(x)g'(x). But what happens when you encounter a product of three or more functions? This article will delve into the intricacies of extending the product rule to handle such scenarios, providing a clear, step-by-step explanation along with illustrative examples. We’ll also explore the underlying mathematical principles and address frequently asked questions.

Understanding the Basic Product Rule

Before tackling the three-function case, let's refresh our understanding of the product rule for two functions. Suppose we have a function h(x) defined as the product of two differentiable functions, f(x) and g(x):

h(x) = f(x)g(x)

The product rule states that the derivative of h(x) is:

h'(x) = f'(x)g(x) + f(x)g'(x)

This means we find the derivative of the first function, multiply it by the second function, and then add the result to the first function multiplied by the derivative of the second function.

Extending the Product Rule: The Case of Three Functions

Now, let's consider a function involving the product of three differentiable functions:

h(x) = f(x)g(x)k(x)

To find the derivative h'(x), we can apply the product rule iteratively. We can treat f(x)g(x) as a single function, say p(x), where p(x) = f(x)g(x). Then, h(x) becomes:

h(x) = p(x)k(x)

Applying the product rule for two functions:

h'(x) = p'(x)k(x) + p(x)k'(x)

Now, we need to find p'(x). Since p(x) = f(x)g(x), we apply the product rule again:

p'(x) = f'(x)g(x) + f(x)g'(x)

Substituting this back into the expression for h'(x):

h'(x) = [f'(x)g(x) + f(x)g'(x)]k(x) + f(x)g(x)k'(x)

Expanding this, we get the product rule for three functions:

h'(x) = f'(x)g(x)k(x) + f(x)g'(x)k(x) + f(x)g(x)k'(x)

Notice a pattern emerging. For each term, we differentiate one function and leave the others unchanged. This pattern generalizes beautifully to more functions, as we'll see shortly.

A General Formula for n Functions

Let's generalize the product rule to a product of n differentiable functions:

h(x) = f₁(x)f₂(x)f₃(x)...fₙ(x)

The derivative h'(x) is given by the sum of n terms, where each term involves the derivative of one of the functions and the product of the remaining functions:

h'(x) = f₁'(x)f₂(x)f₃(x)...fₙ(x) + f₁(x)f₂'(x)f₃(x)...fₙ(x) + f₁(x)f₂(x)f₃'(x)...fₙ(x) + ... + f₁(x)f₂(x)f₃(x)...fₙ'(x)

This formula can be expressed more concisely using summation notation:

h'(x) = Σᵢ [fᵢ'(x) Πⱼ≠ᵢ fⱼ(x)]

where the summation is over i from 1 to n, and the product is over j from 1 to n, excluding j=i. This notation simply encapsulates the same concept: differentiate one function at a time, and multiply by the rest.

Illustrative Examples

Let's illustrate the application of the three-function product rule with some examples.

Example 1:

Let h(x) = x²sin(x)eˣ. Find h'(x).

Here, f(x) = x², g(x) = sin(x), and k(x) = eˣ. Their derivatives are:

f'(x) = 2x g'(x) = cos(x) k'(x) = eˣ

Applying the three-function product rule:

h'(x) = (2x)(sin(x))(eˣ) + (x²)(cos(x))(eˣ) + (x²)(sin(x))(eˣ)

h'(x) = 2xeˣsin(x) + x²eˣcos(x) + x²eˣsin(x)

Example 2:

Let h(x) = (x+1)(x²-2x+1)(x³). Find h'(x).

Here, f(x) = x+1, g(x) = x²-2x+1, and k(x) = x³.

f'(x) = 1 g'(x) = 2x - 2 k'(x) = 3x²

Applying the rule:

h'(x) = (1)(x²-2x+1)(x³) + (x+1)(2x-2)(x³) + (x+1)(x²-2x+1)(3x²)

h'(x) = x⁵ - 2x⁴ + x³ + 2x⁴ - 2x³ + 2x⁴ - 2x³ + 3x⁵ - 6x⁴ + 3x³

h'(x) = 4x⁵ - 2x⁴ + 2x³

The Underlying Mathematical Principle: Iterative Application of the Two-Function Rule

The key to understanding the generalized product rule lies in its iterative nature. It's not a separate, magical formula, but a consequence of repeatedly applying the basic two-function product rule. This iterative process reveals a systematic and predictable pattern, leading to the generalized formula. This recursive application is a powerful technique in many areas of mathematics, emphasizing the interconnectedness of seemingly different concepts.

Dealing with More Than Three Functions

The extension to four or more functions follows the same logic. Each additional function adds another term to the derivative, with each term involving the derivative of one function and the product of the remaining functions. The pattern remains consistent, making it straightforward, albeit potentially tedious, to apply the rule for any number of functions. The summation notation provided earlier is the most efficient way to represent this for a general 'n' number of functions.

Frequently Asked Questions (FAQ)

Q: Is there a shortcut for the product rule with many functions?

A: While there isn't a significantly shorter "shortcut" formula beyond the summation notation, understanding the pattern allows for more efficient calculation. Organize your work systematically to avoid errors. The most efficient approach remains understanding the underlying iterative application of the two-function product rule.

Q: What if one of the functions is not differentiable?

A: The product rule only applies if all functions involved are differentiable at the point of interest. If one of the functions is not differentiable at a particular point, the product rule cannot be used at that point.

Q: Can I use logarithmic differentiation to simplify the problem?

A: Yes, logarithmic differentiation provides an elegant alternative method, particularly helpful when dealing with a product of many functions or complex functions. Taking the natural logarithm of both sides before differentiating simplifies the process by transforming the product into a sum of logarithms, making differentiation significantly easier.

Conclusion

The product rule, initially presented for two functions, gracefully extends to accommodate any number of functions through iterative application. While the expressions can become lengthy, the underlying principle remains consistent and intuitive. By understanding the iterative nature and applying the generalized formula or using logarithmic differentiation, we can efficiently compute the derivatives of products involving any number of differentiable functions. Mastering this concept strengthens your foundation in calculus and opens doors to more advanced applications. Remember to always practice and work through various examples to solidify your understanding.