Problem 32 - Technical Mathematics with Calculus

Question

Evaluate each expression. $$f^{\prime}(2) \text { where } f(x)=e^{\sin (\pi x / 2)}$$

Step-by-Step Solution

Answer

\(f'(2) = 0\)

Step 1: Identify the function and the point of differentiation

First, identify the function to be differentiated, which is given as \(f(x) = e^{\frac{\text{sin}(\text{pi} x)}{2}}\). Also, note that we need to find the derivative at the point \(x = 2\).

Step 2: Differentiate the function

We apply the chain rule to differentiate the function. Let \(u(x) = \text{sin} \big(\frac{\text{pi} x}{2}\big)\). Then the derivative of \(f(x) = e^{u}\) with respect to \(x\) is \(f'(x) = e^{u} \times u'(x)\), where \(u'(x) = \frac{d}{dx} \text{sin} \big(\frac{\text{pi} x}{2}\big)\). To find \(u'(x)\), differentiate \(u(x)\) with respect to \(x\) using the chain rule: \(u'(x)=\frac{\text{pi}}{2} \times \text{cos} \big(\frac{\text{pi} x}{2}\big)\).

Step 3: Evaluate the derivative at the point x=2

Substitute \(x=2\) into the derivative. We get \(f'(2) = e^{\text{sin}(\text{pi}\times2 / 2)} \times \frac{\text{pi}}{2} \times \text{cos} (\text{pi}\times2 / 2)\). Since \(\text{sin}(\text{pi}) = 0\) and \(\text{cos}(\text{pi}) = -1\), the expression simplifies to \(f'(2) = 0\), because the cosine factor becomes \(\text{cos} (\text{pi}) = -1\), but any number multiplied by zero remains zero.

Key Concepts

Derivative of Exponential Function

Understanding the derivative of an exponential function is essential for solving calculus problems involving growth and decay processes, among other applications. The general form of an exponential function is \( e^{x} \), where \( e \) is Euler's number, approximately equal to 2.71828.

When differentiating \( e^{x} \) with respect to \( x \), the derivative is simply \( e^{x} \). However, when the exponent is a more complex function of \( x \)—let's denote it as \( u(x) \)—we must employ the chain rule. The chain rule states that the derivative of \( e^{u(x)} \) with respect to \( x \) is \( e^{u(x)} \) multiplied by the derivative of \( u(x) \) with respect to \( x \), or \( e^{u(x)} \cdot \frac{du}{dx} \).

This concept is crucial when tackling our exercise, where the exponential function has a trigonometric function as its exponent. By applying the chain rule, we obtain the derivative efficiently and properly, thus setting the stage for evaluating the derivative at a specific value of \( x \).

Trigonometric Functions Differentiation

Differentiation of trigonometric functions is another important concept in calculus. The derivatives of the basic trigonometric functions—sine, cosine, and tangent—are as follows:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).
• For the derivative of \( \tan(x) \) it's \( \sec^{2}(x) \).

When differentiating trigonometric functions, it's also important to consider any coefficients or factors that may modify \( x \). In such cases, the chain rule comes into play again.

For example, in our exercise, we differentiate \( \sin(\frac{\pi x}{2}) \), which involves a \( \frac{\pi}{2} \) coefficient. The chain rule tells us to first differentiate \( \sin(x) \) as usual, yielding \( \cos(x) \), and then multiply by the derivative of the inner function \( \frac{\pi x}{2} \) with respect to \( x \)—in this case, \( \frac{\pi}{2} \). As a result, the outer derivative of our trigonometric function is \( \frac{\pi}{2} \cos(\frac{\pi x}{2}) \). This step is crucial before evaluating the derivative at a specific point.

Evaluating Derivatives at a Point

Lastly, we discuss the concept of evaluating derivatives at a point, a common task in calculus that tests our understanding of the differentiation rules. Once we find the derivative of a function, we often need to determine the rate of change at a particular value of \( x \).

To accomplish this, we substitute the value of \( x \) into our derivative formula. Evaluations like these are straightforward but require attention to detail, as making an incorrect substitution or simplification can lead to errors.

In our exercise, once we have the derivative, we find \( f'(2) \) by substituting \( x = 2 \) into our derivative formula. As we perform the substitution, we must remember to consider the trigonometric values of \( \sin(\pi) \) and \( \cos(\pi) \) accurately. Since \( \sin(\pi) = 0 \) and \( \cos(\pi) = -1 \) but any term multiplied by zero leads to zero, we reach the conclusion that \( f'(2) = 0 \). This process encapsulates the application of differentiation rules and the chain rule, and demonstrates how to evaluate the rate of change at a specific point in the domain of a function.