Problem 212 - Calculus Volume 1

Question

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

The second derivative is \( \sec x \tan^2 x + \sec^3 x + 2 \csc^2 x \cot x \).

Step 1: Differentiate the function to find the first derivative

Given the function is \( y = \sec x + \cot x \). We need to find the first derivative \( \frac{dy}{dx} \). First, recall the derivatives:- The derivative of \( \sec x \) is \( \sec x \tan x \).- The derivative of \( \cot x \) is \( -\csc^2 x \).Therefore, the first derivative is:\[ \frac{dy}{dx} = \sec x \tan x - \csc^2 x \]

Step 2: Differentiate the first derivative to find the second derivative

We now compute the second derivative \( \frac{d^2y}{dx^2} \) by differentiating \( \frac{dy}{dx} = \sec x \tan x - \csc^2 x \).Use the product rule for \( \sec x \tan x \):- If \( u = \sec x \) and \( v = \tan x \), then \( (uv)' = u'v + uv' \).- \( u' = \sec x \tan x \) and \( v' = \sec^2 x \).Thus, \( (\sec x \tan x)' = \sec x \tan^2 x + \sec^3 x \).For \( -\csc^2 x \):- The derivative is \( -2 \csc^2 x \cot x \).Combine these:\[ \frac{d^2y}{dx^2} = \sec x \tan^2 x + \sec^3 x + 2 \csc^2 x \cot x \]

Step 3: Simplify the second derivative expression

We need to simplify the expression for the second derivative:\[ \frac{d^2y}{dx^2} = \sec x \tan^2 x + \sec^3 x + 2 \csc^2 x \cot x \]Notice this expression does not simplify much further directly, as each component has a separate trigonometric basis (secant/tangent versus cosecant/cotangent). Therefore, this is the final step for simplification of the second derivative.

Key Concepts

Higher-order derivative

In calculus, the idea of a higher-order derivative is an extension of the concept of a derivative. When you differentiate a function, you obtain its first derivative, which describes the rate of change of the function. The second derivative, or higher-order derivatives, describe the rate of change of the rate of change, adding depth to your understanding of how a function behaves.
Higher-order derivatives are particularly useful in physics and engineering, where they can describe things like acceleration, concavity, and more. To find the second derivative, for example, you typically start by finding the first derivative, then derive yet again.
For the function given in the exercise, the progression from the first to the second derivative involves standard differentiation techniques, as discussed further in the next sections.

Trigonometric functions

Trigonometric functions, such as secant (\( \sec x \)) and cotangent (\( \cot x \)), play a key role in calculus due to their repetitive and oscillating nature, which are ideal for modeling waves and periodic phenomena.
Understanding the derivatives of these trigonometric functions is essential in solving calculus problems.Recall:

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

These derivatives are derived from the fundamental properties of trigonometric functions and their limits.

Differentiation rules

Differentiation rules in calculus are formulas that help you find derivatives of functions more efficiently. These rules form the backbone of calculus operations and are invaluable for solving complex functions quickly.
For basic trigonometric functions, there are set rules, as previously mentioned, which allow for the smooth computation of derivatives. When dealing with sums or differences of functions, the sum and difference rules allow each part to be differentiated separately.
The differentiation operation follows linearity, so \( rac{d}{dx} [f(x) + g(x)] = f'(x) + g'(x) \), where \( f'(x) \) and \( g'(x) \) are individual derivatives of \( f(x) \) and \( g(x) \), respectively.

Product rule

The product rule is a fundamental differentiation rule used when differentiating products of two functions. This rule states that the derivative of a product is not simply the product of the derivatives.
Instead, if you have a function \( y = uv \), where \( u \) and \( v \) are functions of \( x \), the product rule formula is:\[ (uv)' = u'v + uv' \]Consider the example from the exercise, where \( y = \sec x \tan x \). Using the product rule:

• Let \( u = \sec x \) and \( v = \tan x \).
• The derivative \( u' = \sec x \tan x \).
• The derivative \( v' = \sec^2 x \).

Apply the rule: \( (\sec x \tan x)' = \sec x \tan^2 x + \sec^3 x \), beautifully demonstrating how the product rule works with trigonometric functions.
This step is crucial for finding derivatives of expressions where functions are multiplied together, as it ensures accurate results by considering both terms together as a linked pair.