Problem 20 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

The second derivative is \( \frac{d^2 y}{dx^2} = \csc^3 x - \csc x \cot^2 x \).

Step 1: Understand the Problem

We need to find the second derivative of the function \( y = \csc x \). This requires us to first find the first derivative \( \frac{dy}{dx} \) and then differentiate that result once more to obtain the second derivative \( \frac{d^2y}{dx^2} \).

Step 2: Find the First Derivative

Recall that \( y = \csc x \) can be expressed as \( y = \frac{1}{\sin x}\). Hence, the first derivative is \( \frac{dy}{dx} = -\csc x \cot x \). This is obtained using the chain rule and the derivative of \( \csc x \).

Step 3: Differentiate the First Derivative

We next differentiate \( \frac{dy}{dx} = -\csc x \cot x \) to find the second derivative. Use the product rule: \[ \frac{d}{dx}(-\csc x \cot x) = -\csc x \frac{d}{dx}(\cot x) + (-\cot x) \frac{d}{dx}(\csc x) \].

Step 4: Apply Derivatives

Calculate each derivative: - The derivative of \( \cot x \) is \( -\csc^2 x \).- The derivative of \( \csc x \) is \( -\csc x \cot x \).Substitute these into the product rule equation.

Step 5: Simplify the Expression

Substitute the derivatives found in Step 4: \[ \frac{d}{dx}(-\csc x \cot x) = -\csc x (-\csc^2 x) + \csc x \cot^2 x \]This simplifies to:\[ \csc^3 x - \csc x \cot^2 x \].

Step 6: Finalize the Solution

The second derivative \( \frac{d^2 y}{dx^2} \) simplifies from its components to: \[ \frac{d^2 y}{dx^2} = \csc^3 x - \csc x \cot^2 x \].

Key Concepts

Cosecant Function

The cosecant function, denoted as \( \csc x \), is one of the lesser-known trigonometric functions. It is the reciprocal of the sine function. In mathematical terms, this means \( \csc x = \frac{1}{\sin x} \). The cosecant function is defined for all values of \( x \) where sine is not zero. This restriction excludes integer multiples of \( \pi \), where \( \sin x = 0 \). Understanding how \( \csc x \) behaves is foundational for working with its derivatives.

• The range of \( \csc x \) is from negative infinity to -1 and from 1 to positive infinity.
• In a graph, \( \csc x \) has discontinuities (vertical asymptotes) where sine is zero.

Working with \( \csc x \) often involves transforming it into its sine form to make derivative calculations easier. This transformation utilizes its reciprocal identity.

Product Rule

The product rule is an essential tool in calculus for finding the derivative of a product of two functions. When you have a function \( g(x) \) and another function \( h(x) \), the product rule states:\[ \frac{d}{dx}[g(x) \cdot h(x)] = g(x) \cdot \frac{d}{dx}[h(x)] + h(x) \cdot \frac{d}{dx}[g(x)] \]
This rule is particularly useful when dealing with trigonometric derivatives, as in finding the second derivative of \( y = -\csc x \cot x \). Let's see how it is applied:

• Begin by identifying \( g(x) = -\csc x \) and \( h(x) = \cot x \) in our example.
• Then, differentiate \( h(x) = \cot x \) and \( g(x) = -\csc x \) separately.
• Apply these derivatives in the product rule formula to find the overall derivative of the product.

This approach helps breakdown complex expressions into more manageable parts.

Chain Rule

The chain rule is another fundamental technique in calculus used when differentiating composite functions, meaning one function inside another. If a function \( f(x) \) can be defined as \( f(g(x)) \), then its derivative is expressed using the chain rule as:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]The chain rule is critical for trigonometric derivatives, like when transforming \( \csc x = \frac{1}{\sin x} \).

• Identify the outer function, which, in the case of cosecant, is \( \frac{1}{x} \).
• Recognize the inner function as \( \sin x \).
• Differentiate each part and apply the chain rule to find the derivative of \( \csc x \).

This step-by-step differentiation yields \( -\csc x \cot x \), the derivative of \( \csc x \). This makes the chain rule indispensable for working with nested functions in derivatives.

Trigonometric Derivatives

Trigonometric derivatives are derivatives concerning trigonometric functions such as sine, cosine, tangent, and their reciprocals, like cosecant. These derivatives allow us to analyze how trigonometric functions change or respond to input changes.For the function \( \csc x \):

• The first derivative is \( \frac{d}{dx}[ \csc x] = -\csc x \cot x \).
• Further differentiation involves knowing the derivatives of each trigonometric component.

Dealing with the second derivative of \( \csc x \) requires applying the product rule after finding the first derivative using trigonometric identities.Understanding trigonometric derivatives is vital for applications in physics and engineering, where wave patterns and oscillations are common. Mastery of these derivatives provides a strong foundation for tackling more complex calculus problems involving trigonometric functions.