Problem 212 - Calculus Volume 1

Question

For the following exercises, find the requested higher-order derivative for the given functions. $$\frac{d^{2} y}{d x^{2}}\text { of } y=\sec x+\cot x$$

Step-by-Step Solution

Answer

\( \frac{d^2y}{dx^2} = \sec^3 x + \sec x \tan^2 x + 2\csc^2 x \cot x \)

Step 1: Differentiation of Each Term

Start by finding the first derivative of the function \( y = \sec x + \cot x \). Differentiate each term separately.For \( \frac{d}{dx}(\sec x) \):\[ \frac{d}{dx}(\sec x) = \sec x \tan x \]For \( \frac{d}{dx}(\cot x) \):\[ \frac{d}{dx}(\cot x) = -\csc^2 x \]Thus, the first derivative is:\[ \frac{dy}{dx} = \sec x \tan x - \csc^2 x \]

Step 2: Derive the First Derivative Again

Now, find the second derivative, i.e., \( \frac{d^2y}{dx^2} \), by differentiating \( \frac{dy}{dx} = \sec x \tan x - \csc^2 x \) again.Differentiating \( \sec x \tan x \):Using the product rule:\[ \frac{d}{dx}(\sec x \tan x) = \sec x \frac{d}{dx}(\tan x) + \tan x \frac{d}{dx}(\sec x) = \sec x \sec^2 x + \tan x \sec x \tan x = \sec^3 x + \sec x \tan^2 x \]Differentiating \( -\csc^2 x \):\[ \frac{d}{dx}(-\csc^2 x) = -(-2\csc x \cdot \csc x \cot x) = 2\csc^2 x \cot x \]Thus, the second derivative is:\[ \frac{d^2y}{dx^2} = \sec^3 x + \sec x \tan^2 x + 2\csc^2 x \cot x \]

Key Concepts

Differentiation

Differentiation is a fundamental concept of calculus that deals with finding the rate at which a function changes. When you differentiate a function, you are essentially finding its slope or the rate of change related to the function's variable. This is especially important when you need to understand how a function behaves as its input changes.

When differentiating, you're typically dealing with functions that have one or more terms. Each term needs to be considered individually. For example, in the exercise, the function is given as \( y = \sec x + \cot x \). Differentiating each term separately leads us to the first derivative. For \( \sec x \), the derivative is \( \sec x \tan x \). Likewise, the derivative of \( \cot x \) is \( -\csc^2 x \).

These derivatives help in determining how each component of the function is changing and plays a critical role when finding higher-order derivatives, which are derivatives of derivatives. By finding the first derivative, you set up the framework needed to proceed with calculating the second derivative, further examining the behavior of the function.

Product Rule

The product rule is an essential technique used in differentiation when a function is the product of two or more smaller functions. It provides a way to differentiate products of functions, ensuring that each component of the product is differentiated properly. The rule states that for any two differentiable functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x) \cdot v(x) \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

In the exercise, the product rule is used while differentiating \( \sec x \tan x \). First, differentiate \( \sec x \) and multiply by \( \tan x \), then \( \tan x \) is differentiated and multiplied by \( \sec x \). This can be visualized as following these steps:

• Differentiate \( \tan x \) to get \( \sec^2 x \), and multiply by \( \sec x \) from the first term.
• Then, \( \sec x \) is multiplied by the derivative of \( \tan x \), which results in \( \sec x \tan^2 x \).
• Finally, summing these products gives \( \sec^3 x + \sec x \tan^2 x \).

Understanding and applying the product rule correctly is crucial for calculating derivatives of more complex functions that involve products.

Trigonometric Functions

Trigonometric functions such as \( \sec x \), \( \tan x \), \( \cot x \), and \( \csc x \) frequently appear in calculus problems. These functions have specific derivatives which are fundamental to know. For instance, the derivative of \( \sec x \) is \( \sec x \tan x \), while the derivative of \( \cot x \) is \( -\csc^2 x \).

These derivatives are directly used when calculating higher-order derivatives and are critical when analyzing functions involving angles, oscillations, and waves. In trigonometric calculus, knowing how to differentiate these functions is a stepping stone for more advanced problems.

Trigonometric functions also have unique identities and properties that are often employed during differentiation. For example, understanding that \( \tan x = \frac{\sin x}{\cos x} \) helps when simplifying derivatives or solving equations involving trigonometric terms. Recognizing these patterns and properties allows you to calculate derivatives efficiently, especially when combined with rules like the product rule.