Question

True or False If \(u\) and \(v\) are differentiable functions, then \(d(u v)=d u d v .\) Justify your answer.

Short answer

The given statement is False.

Step-by-step solution

Statement Analysis

First, analyze the given statement 'If \(u\) and \(v\) are differentiable functions, then \(d(uv)=du \cdot dv\).' Here, \(d(uv)\) represents the differential of the product of functions \(u\) and \(v\), and \(du \cdot dv\) represents the product of their differentials.

Apply Product Rule for Differentiation

Recall the product rule for differentiation which states if two functions \(u\) and \(v\) are differentiable then their product's differential is given by \(d(uv) = u \cdot dv + v \cdot du\).

Compare the Result with the Statement

Now, compare this result with the given statement. You will notice that the given statement says that \(d(uv)=du \cdot dv\), which does not match our result obtained from the product rule of differentiation.

Key concepts

Differential Calculus

Differential calculus is a subfield of calculus concerned with the concept of a derivative, which represents an instantaneous rate of change. It is incredibly useful in various fields such as physics, engineering, and economics as it allows the understanding of how a function's value changes as its inputs change.

When we look at the rate at which one quantity changes with respect to another, we are dealing with differential calculus. For instance, if a car's position changes over time, differential calculus can help us to understand its speed at any given moment. Calculus educators emphasize the importance of understanding the derivative not just as a formula, but as a fundamental concept that describes change.

An example of a differential operation is finding the derivative of a product of two functions, which is not simply the product of the derivatives of each function, but instead uses the product rule, a cornerstone in differential calculus.

Differentiable Functions

A function is differentiable if it has a derivative at every point in its domain. This means that the function is smooth and continuous, without any sharp edges or breaks. Differentiable functions are important in calculus because they ensure that derivatives can be applied across the entire function without encountering undefined behavior.

In the context of the original exercise, both functions, represented by the variables 'u' and 'v', are assumed to be differentiable. However, it is critical to highlight that the derivative of the product of two differentiable functions is not simply obtained by multiplying their derivatives directly. This common misconception can lead to errors in calculus problem-solving. The correct method involves applying the product rule, as demonstrated in the step-by-step solution.

Calculus Education

Effective calculus education hinges on students developing a strong conceptual understanding of derivatives and their applications. The product rule is one such application that illustrates the nuanced behavior of derivatives when applied to the product of functions. Students must recognize that rules like this cannot be assumed intuitively and must be learned and applied with precision.

Providing clear examples and emphasizing the non-trivial nature of rules like the product rule will help students avoid errors and develop deeper mathematical skills. As seen with the exercise on the product of two differentiable functions, teaching why the result is not simply the product of the derivatives and justifying each step is crucial for true comprehension. Enhancing calculus education means embracing these deeper explanations and encouraging learners to question and understand every detail.