Problem 13 - Single Variable Calculus

Question

Find the differential of the function at the indicated number. $$ f(x)=(1+2 \cos x)^{1 / 2} ; \quad x=\frac{\pi}{2} $$

Step-by-Step Solution

Answer

The differential of the function \(f(x) = x \tan x\) at \(x = \frac{\pi}{4}\) is \(f'\left(\frac{\pi}{4}\right) = 1 + \frac{\pi}{2}\).

Step 1: Differentiate the function with respect to x.

We have the function \(f(x) = x \tan x\). To differentiate this, we will use the product rule. The product rule states that if \(f(x) = g(x)h(x)\), then \(f'(x) = g'(x)h(x) + g(x) h'(x)\), where \(g'(x)\) and \(h'(x)\) are the derivatives of \(g(x)\) and \(h(x)\) respectively. In this case, we have \(g(x) = x\) and \(h(x) = \tan x\).

Step 2: Differentiate \(g(x)\) and \(h(x)\).

First differentiate \(g(x) = x\) with respect to \(x\), which gives \(g'(x) = 1\). Next, differentiate \(h(x) = \tan x\) with respect to \(x\), which gives \(h'(x) = \sec^2 x\), as the derivative of \(\tan x\) is \(\sec^2 x\).

Step 3: Apply the product rule.

Now that we have \(g'(x) = 1\) and \(h'(x) = \sec^2 x\), we can use the product rule to find the derivative of \(f(x) = x \tan x\). Using the formula mentioned earlier, we get: \(f'(x) = g'(x)h(x) + g(x) h'(x) = (1)(\tan x) + (x)(\sec^2 x) = \tan x + x \sec^2 x\).

Step 4: Evaluate the derivative at the given point.

Now we need to find the differential of the function at \(x = \frac{\pi}{4}\), so we need to evaluate \(f'(x)\) at this point: \(f'\left(\frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) + \frac{\pi}{4} \sec^2\left(\frac{\pi}{4}\right)\). Since \(\tan\left(\frac{\pi}{4}\right) = 1\) and \(\sec\left(\frac{\pi}{4}\right) = \sqrt{2}\), we can substitute these values into the equation: \(f'\left(\frac{\pi}{4}\right) = 1 + \frac{\pi}{4} (2) = 1 + \frac{\pi}{2}\). So, the differential of the function at \(x = \frac{\pi}{4}\) is \(f'\left(\frac{\pi}{4}\right) = 1 + \frac{\pi}{2}\).

Key Concepts

Product Rule

The Product Rule is an essential concept in differential calculus. It is used to find the derivative of a product of two functions. When you have a function like \( f(x) = g(x)h(x) \), and you need to differentiate it, the product rule comes into play.
To apply the product rule, remember this simple formula:

• \( f'(x) = g'(x)h(x) + g(x)h'(x) \)

This formula implies that you need the derivatives of \( g(x) \) and \( h(x) \) respectively. Then, you multiply the derivative of the first function by the second function as is, and vice-versa, and add those results together.

In our example:

• Let \( g(x) = x \) and \( h(x) = \tan x \).
• The derivative \( g'(x) = 1 \).
• The derivative \( h'(x) = \sec^2 x \).

Applying the product rule, we get:

• \( f'(x) = 1(\tan x) + x (\sec^2 x) = \tan x + x \sec^2 x \)

This handy rule simplifies the process when dealing with the multiplication of two functions.

Trigonometric Functions

Trigonometric functions form the backbone of many mathematical applications. Functions such as \( \sin x \), \( \cos x \), and \( \tan x \) are commonly used. These functions describe relationships in triangles and have various periodic properties.

In calculus, differentiating these functions involves knowing their respective derivates:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).
• The derivative of \( \tan x \) is \( \sec^2 x \).

In our exercise, we used \( \tan x \), whose derivative \( \sec^2 x \) is essential for applying the product rule effectively. Understanding these derivatives helps solve problems quicker and more efficiently.

Additionally, knowing values such as \( \tan(\frac{\pi}{4}) = 1 \) and \( \sec(\frac{\pi}{4}) = \sqrt{2} \) allows for straightforward substitution in evaluations at specific points.

Derivative Evaluation

Derivative evaluation involves both finding and calculating derivatives at specific points, which is a common task in calculus. This entails differentiating a function and then substituting a specific value for \( x \).

Let's use the derivative from the exercise again:

• \( f'(x) = \tan x + x \sec^2 x \)

To perform derivative evaluation for \( x = \frac{\pi}{4} \):

• Substitute: \( f'(\frac{\pi}{4}) = \tan(\frac{\pi}{4}) + \frac{\pi}{4} \sec^2(\frac{\pi}{4}) \)
• Evaluate using known trigonometric values: \( \tan(\frac{\pi}{4}) = 1 \) and \( \sec^2(\frac{\pi}{4}) = 2 \).
• The result is \( 1 + \frac{\pi}{2} \).

By understanding how to evaluate derivatives at particular points, you gain the ability to predict the rate of change in a function accurately at those points. This is incredibly useful for understanding behavior and trends within different functions.