Problem 36 - Calculus Single Variable

Question

Calculate the requested derivative. \(H^{(2)}(\pi / 4)\) where \(H(x)=x \tan (x)\)

Step-by-Step Solution

Answer

The second derivative is \( H^{(2)}(\pi/4) = 4 + \pi \).

Step 1: Identify the function

The function given is \( H(x) = x \tan(x) \). We are asked to find the second derivative of this function, \( H^{(2)}(x) \), and evaluate it at \( x = \pi/4 \).

Step 2: Find the first derivative

To find the first derivative, \( H^{(1)}(x) \), use the product rule: \( (uv)' = u'v + uv' \). Let \( u = x \) and \( v = \tan(x) \). Then \( u' = 1 \) and \( v' = \sec^2(x) \). The first derivative is: \[H^{(1)}(x) = x \cdot \sec^2(x) + \tan(x) \cdot 1 = x \sec^2(x) + \tan(x).\]

Step 3: Find the second derivative

Differentiate \( H^{(1)}(x) = x \sec^2(x) + \tan(x) \) again. This involves the product rule for \( x \sec^2(x) \) and the simple derivative rule for \( \tan(x) \):\( (x \sec^2(x))' = 1 \cdot \sec^2(x) + x \cdot 2\sec^2(x)\tan(x) = \sec^2(x) + 2x\sec^2(x)\tan(x) \).For \( \tan(x) \), the derivative is \( \sec^2(x) \).Thus, in total:\[H^{(2)}(x) = \sec^2(x) + 2x\sec^2(x)\tan(x) + \sec^2(x) = 2\sec^2(x) + 2x\sec^2(x)\tan(x).\]

Step 4: Evaluate the second derivative at \( x = \pi/4 \)

Substitute \( x = \pi/4 \) into the expression for \( H^{(2)}(x) \): \[H^{(2)}\left(\frac{\pi}{4}\right) = 2\sec^2\left(\frac{\pi}{4}\right) + 2\left(\frac{\pi}{4}\right)\sec^2\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{4}\right).\]Since \( \tan(\pi/4) = 1 \) and \( \sec(\pi/4) = \sqrt{2} \), we have \( \sec^2(\pi/4) = 2 \). Plug these values in:\[H^{(2)}\left(\frac{\pi}{4}\right) = 2(2) + 2\left(\frac{\pi}{4}\right)(2)(1)\]\[= 4 + \pi.\]

Step 5: Final Calculation

The above steps simplify the expression to calculate the value of the second derivative at \( x = \pi/4 \). Therefore, the result is:\[H^{(2)}\left(\frac{\pi}{4}\right) = 4 + \pi.\]

Key Concepts

Product Rule

The product rule is an essential tool in calculus used when differentiating products of two functions. The rule states that if you have a function that is a product of two other functions, say \( u(x) \) and \( v(x) \), then the derivative of their product, denoted as \((uv)'\), can be found using the formula:

• \((uv)' = u'v + uv'\)

Applying this rule helps break down complex derivatives into manageable parts.
In our exercise, the function \( H(x)=x \tan(x) \) is indeed a product of two functions: \( u = x \) and \( v = \tan(x) \). To find the first derivative, \( H^{(1)}(x) \), you start by differentiating each component separately.
For \( u = x \), the derivative \( u' \) is straightforward and equal to 1. For \( v = \tan(x) \), the derivative \( v' \) is \( \sec^2(x) \) because the derivative of \( \tan(x) \) is \( \sec^2(x) \).
Using the product rule, we combine these derivatives to find the first derivative:

• \( H^{(1)}(x) = x \cdot \sec^2(x) + \tan(x) \cdot 1 = x \sec^2(x) + \tan(x) \)

Understanding the product rule is crucial for correctly evaluating derivatives of products and is widely applicable in various fields of mathematics and physics.

Trigonometric Functions

Trigonometric functions are a cornerstone in calculus, especially when dealing with derivatives. Understanding these functions, such as sine, cosine, tangent, and their reciprocals, is essential for tackling calculus problems.
In this exercise, we focus on \( \tan(x) \) and \( \sec(x) \). The tangent function \( \tan(x) \) is defined as the ratio of the sine and cosine functions. It has interesting properties, one of which is evident in its derivative:

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

The secant function \( \sec(x) \) is the reciprocal of the cosine function, and the derivative properties of \( \tan(x) \) play a significant role in solving our derivative exercise.
When evaluating the function \( H^{(2)}(x) \) at \( x = \pi/4 \), it is useful to know specific values, such as \( \tan(\pi/4) = 1 \) and \( \sec(\pi/4) = \sqrt{2} \), leading to \( \sec^2(\pi/4) = 2 \).
These specific values simplify calculations immensely and highlight the importance of understanding trigonometric functions and their derivatives thoroughly.

Second Derivative

Finding the second derivative is a step further in understanding the behavior of functions. It describes how the rate of change of a function's rate of change is changing; essentially, it provides insights into the function's concavity and points of inflection.
To find the second derivative of a function like \( H(x) = x \tan(x) \), you first determine the first derivative using rules like the product rule. Once you have the first derivative, in this case, \( H^{(1)}(x) = x \sec^2(x) + \tan(x) \), you differentiate it again.
This process involves applying the product rule once more for parts like \( x \sec^2(x) \) and processing \( \tan(x) \) as well. Here's how it looks:

• \( (x \sec^2(x))' = 1 \cdot \sec^2(x) + x \cdot 2\sec^2(x)\tan(x) = \sec^2(x) + 2x\sec^2(x)\tan(x) \)
• The derivative of \( \tan(x) \) is \( \sec^2(x) \), so add \( \sec^2(x) \) to the result.

After combining these results, we find the second derivative:

• \( H^{(2)}(x) = 2\sec^2(x) + 2x\sec^2(x)\tan(x) \)

Finally, you substitute \( x = \pi/4 \) to get \( H^{(2)}(\pi/4) = 4 + \pi \). The second derivative helps in ascertaining how steeply a function is curving or changing, making it invaluable in detailed curve analysis.