Problem 45 - Calculus Early Transcendentals

Question

Consider the following functions and express the relationship between a small change in \(x\) and the corresponding change in \(y\) in the form \(d y=f^{\prime}(x) d x\). \(f(x)=2-a \cos x, a\) constant

Step-by-Step Solution

Answer

Answer: The relationship between a small change in \(x\) and the corresponding change in \(y\) for the function \(f(x) = 2 - a \cos x\) is given by \(dy = a\sin x \, dx\).

Step 1: Find the derivative of the function

To find the derivative \(f'(x)\) of the function \(f(x) = 2 - a \cos x\), we will apply the differentiation rules for constant, cosine, and subtraction functions. Recall that the derivative of a constant is 0, and the derivative of \(\cos x\) is \(-\sin x\). Therefore, we have the following differentiation: $$f'(x) = \frac{d}{dx}(2 - a \cos x) = -a\frac{d}{dx}(\cos x)$$ Now, apply the derivative of the cosine function: $$f'(x) = -a(-\sin x) = a\sin x$$

Step 2: Express the relationship between small changes in \(x\) and \(y\)

Now that we have \(f'(x) = a\sin x\), we can express the small change in \(y\), denoted by \(dy\), in terms of a small change in \(x\), denoted by \(dx\). We use the given formula \(dy = f'(x)dx\): $$dy = a\sin x \, dx$$ So, the relationship between a small change in \(x\) and the corresponding change in \(y\) for the function \(f(x) = 2 - a \cos x\) is given by: $$dy = a\sin x \, dx$$

Key Concepts

Derivative of a Function

The derivative of a function is, at its core, a measure of how a function's output value changes as its input value changes. In formal terms, it represents the rate of change of the function value with respect to its variable. For instance, when dealing with a function like
\[\begin{equation} f(x) = 2 - a \cos x, \end{equation}\]
where \( a \) is a constant, we determine the derivative by separately analyzing each component of the function. First, we note that the derivative of a constant (like '2' in our function) is zero because constants do not change, no matter the input. Next, we look at the cosine component. In calculus, the derivative of \( \cos x \) is \( -\sin x \). Combining these elements, we arrive at the derivative:
\[\begin{equation} f'(x) = a\sin x. \end{equation}\]
Understanding derivatives is essential because they enable us to analyze and predict function behavior, especially concerning changes in variables like time, space, or any quantifiable entity.

Chain Rule

Occasionally, we encounter composite functions, which are essentially functions within functions. The chain rule is a fundamental tool in calculus used to derive the rate of change of such composite functions. It states that if you have a function \(g(x)\) inside another function \(f(u)\), and \(u = g(x)\), the derivative of the compound function \(f(g(x))\) is the derivative of \(f\) with respect to \(u\), multiplied by the derivative of \(g\) with respect to \(x\):
\[\begin{equation} \frac{d}{dx}f(g(x)) = f'(g(x)) \times g'(x). \end{equation}\]
This allows us to differentiate more complex structures by systematically breaking them down into simpler parts. While our exercise doesn't directly involve the chain rule, understanding its mechanics is crucial for tackling advanced differential calculus problems.

Small Changes in Functions

When we talk about small changes in functions, we refer to the concept of differentials. In a practical sense, if we have a small change in \(x\), denoted as \(dx\), and we want to know the corresponding small change in the function's value, \(y\), we would use the function's derivative multiplied by \(dx\). The formula
\[\begin{equation} dy = f'(x)dx \end{equation}\]
is used to represent this relationship. This equation can be interpreted as the linear approximation of the function’s change in a tiny interval around a certain point. Using the solution of our example,
\[\begin{equation} dy = a\sin x \times dx, \end{equation}\]
we can estimate how much \(y\) varies with a minuscule increase or decrease in \(x\). Differentials are incredibly useful in scientific and engineering contexts, where they aid in calculating error margins and in optimizations.

Trigonometric Functions Differentiation

Differentiating trigonometric functions is a key aspect of calculus, especially given their wide application across various fields. Each trigonometric function has its derivative that can be memorized or derived using the unit circle. For instance,
\[\begin{equation} \frac{d}{dx}(\sin x) = \cos x, \end{equation}\]
and as seen in our exercise,
\[\begin{equation} \frac{d}{dx}(\cos x) = -\sin x. \end{equation}\]
Other trigonometric functions such as tangent and secant have their respective derivatives, too. The process requires recognizing the function at hand and applying its specific differentiation rule. Memorizing these basic derivatives forms a foundation enabling you to navigate through more complex differential calculus problems involving trigonometric functions.