Problem 41 - Essential Calculus Early Transcendentals

Question

Find the derivative. Simplify where possible. $$ y=\operatorname{coth}^{-1}(\sec x) $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = \csc x\)

Step 1: Understand the Function and Its Derivative

To find the derivative of the given function, first, identify the outer and inner functions. The function given is the inverse hyperbolic cotangent of the secant of \(x\). We'll use the chain rule to differentiate this composite function. Recall that the derivative of \( y = \operatorname{coth}^{-1}(u) \) with respect to \(u\) is \( \frac{-1}{1-u^2} \).

Step 2: Differentiate the Outer Function

Using the derivative formula from Step 1 for \( y = \operatorname{coth}^{-1}(u) \), where \( u = \sec x \), we have:\[\frac{dy}{du} = -\frac{1}{1-(\sec x)^2}\]

Step 3: Simplify the Derivative of the Outer Function

Recognize that \(1 - (\sec x)^2 = -\tan^2 x\). So, replace it in:\[\frac{dy}{du} = -\frac{1}{-\tan^2 x} = \frac{1}{\tan^2 x}\]

Step 4: Differentiate the Inner Function

The inner function is \( u = \sec x \). The derivative of \( \sec x \) with respect to \( x \) is \( \sec x \tan x \). Thus:\[\frac{du}{dx} = \sec x \cdot \tan x\]

Step 5: Apply the Chain Rule

Now, use the chain rule to combine the results of the outer and inner derivatives. The chain rule states that:\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]Substitute the derivatives found in steps 3 and 4:\[\frac{dy}{dx} = \frac{1}{\tan^2 x} \cdot \sec x \cdot \tan x\]Which simplifies to:\[\frac{dy}{dx} = \frac{\sec x}{\tan x}\]

Step 6: Simplify the Expression Further

Recognize that \( \tan x = \frac{\sin x}{\cos x} \), so \( \frac{1}{\tan x} = \frac{\cos x}{\sin x} = \cot x \). Thus:\[\frac{dy}{dx} = \sec x \cdot \cot x\]Since \( \sec x = \frac{1}{\cos x} \), we have:\[\frac{dy}{dx} = \frac{1}{\cos x} \cdot \frac{\cos x}{\sin x} = \frac{1}{\sin x} = \csc x\]

Key Concepts

Inverse Hyperbolic Function

Inverse hyperbolic functions are the inverses of the standard hyperbolic functions. These functions have similar properties to the inverse trigonometric functions but belong to the hyperbolic sphere. Examples include inverse hyperbolic sine, cosine, and, as in our exercise, cotangent, denoted as \(\operatorname{coth}^{-1}\). Inverse hyperbolic cotangent, specifically, maps real numbers back to their respective hyperbolic cotangent values. It serves as a tool to reverse the effect of hyperbolic cotangent. The main challenge while working with inverse hyperbolic functions arises from identifying how the inner and outer components of a composite function interact, which is crucial when applying differentiation techniques like the chain rule.

Chain Rule

The chain rule is fundamental to calculating derivatives of composite functions. When a function is composed of two or more functions, its derivative is calculated by sequentially applying the rules of differentiation. First, differentiate the outer function while keeping the inner function intact, then multiply by the derivative of the inner function. In symbolic terms, if \(y = f(g(x))\), then the derivative \(\frac{dy}{dx}\) is \(f'(g(x)) \cdot g'(x)\).

• Outer function: Often the main or observable function, to be differentiated first.
• Inner function: Nested within the outer function, affects the derivative of the whole.

In our exercise, the chain rule was utilized by differentiating the inverse hyperbolic cotangent with respect to secant and then multiplying by the derivative of the secant function.

Differentiation Techniques

Differentiation techniques include various methods to find the derivative of a function. These methods involve the power rule, product rule, quotient rule, and chain rule, among others. For complicated expressions, multiple techniques may need to be combined. In this exercise, the composite function \(y = \operatorname{coth}^{-1}(\sec x)\) requires the use of the chain rule. Additionally, knowledge of the derivatives of trigonometric functions such as secant and tangent is applied.
Transformations and simplifications, such as those using trigonometric identities like \(1 - (\sec x)^2 = -\tan^2 x\), can further simplify the process.

• Recognizing trigonometric identities aids in simplifying derivatives.
• Substituting components with these identities streamlines computation.

These techniques collectively provide a streamlined approach to tackle complex calculus problems.