Problem 20 - Calculus Early Transcendentals

Question

Find \(d^{2} y / d x^{2}\) $$ y=\csc x $$

Step-by-Step Solution

Answer

The second derivative is \( \csc^3 x \).

Step 1: Find the First Derivative

To find the second derivative, we first need the first derivative of the function. Given the function \( y = \csc x \), recall that the derivative of \( \csc x \) is \( -\csc x \cot x \). Thus, the first derivative \( \frac{dy}{dx} \) is \( -\csc x \cot x \).

Step 2: Find the Second Derivative

Now, we take the derivative of \( \frac{dy}{dx} = -\csc x \cot x \). Use the product rule for differentiation which is: \((uv)' = u'v + uv'\). Let \( u = -\csc x \) and \( v = \cot x \). Then, \( u' = \csc x \cot x \) and \( v' = -\csc^2 x \).

Step 3: Apply Product Rule

Applying the product rule to \( \frac{d}{dx} (-\csc x \cot x) \):\[ u'v + uv' = (\csc x \cot x)(\cot x) + (-\csc x)(-\csc^2 x) \]This becomes:\[ \csc x \cot^2 x + \csc^3 x \]

Step 4: Simplify the Expression

Combine the terms and simplify:\[ \csc x \cot^2 x + \csc^3 x = \csc x (\cot^2 x + \csc^2 x) \] Since \( \cot^2 x + \csc^2 x = \csc^2 x \), the expression simplifies to:\[ \csc x \cdot \csc^2 x = \csc^3 x \]Thus, the second derivative is \( \frac{d^2y}{dx^2} = \csc^3 x \).

Key Concepts

Trigonometric Derivatives

Trigonometric derivatives play a crucial role in calculus, especially when you work with functions involving angles. Understanding these derivatives is vital to solving problems related to trigonometric functions. One of the essential derivatives to remember is that of the cosecant function, denoted as \( \csc x \).
In calculus, the derivative of \( \csc x \) is \(-\csc x \cot x\). This derivative emerges from applying basic differentiation rules to trigonometric functions.
Here are some key points about why these derivatives are so important:

• Provide tools for understanding the rate of change within triangles and circular functions.
• Crucial for finding slopes of tangent lines and determining concavity.
• Enable the calculation of higher-order derivatives as seen in this exercise.

Anyone working with calculus will frequently encounter these derivatives, so consolidating knowledge of each trigonometric function's derivative ensures success in tackling complex calculus exercises.

Product Rule

The product rule is a fundamental differentiation rule used when a function is the product of two or more functions. It allows the calculation of derivatives of products of two functions.
In our exercise, when finding the second derivative of \( \csc x \), the first derivative \(-\csc x \cot x\) is a product of two functions: \(-\csc x\) and \(\cot x\). Here's how the product rule works:

• Consider two functions: \( u(x) \) and \( v(x) \).
• The product rule is expressed as \( (uv)' = u'v + uv' \).
• This means the derivative of the product is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

If applied correctly, the product rule simplifies calculus operations and ensures accurate solutions when dealing with multiplying functions.

Csc Function

The cosecant function, denoted as \( \csc x \), is one of the six fundamental trigonometric functions often used in calculus and trigonometry.
The cosecant function is defined as \( \csc x = \frac{1}{\sin x} \).
This definition immediately shows that \( \csc x \) is undefined when \( x = n\pi \), where \( n \) is an integer. It is crucial to keep this in mind when dealing with problems involving \( \csc x \) to avoid indeterminate values.
Some important properties of the cosecant function include:

• Because \( \csc x \) is the reciprocal of \( \sin x \), its graphs and behavior are influenced by those of the sine function.
• Due to its relationship with the sine function, \( \csc x \) helps in solving trigonometric identities and integral calculus problems.
• The behavior of \( \csc x \) over its domain requires careful analysis around points where it is undefined.

Understanding \( \csc x \) will enhance your ability to handle trigonometric functions in various mathematical contexts, making calculus operations involving this function more intuitive.