Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\ln (\operatorname{coth} x) $$

Step-by-Step Solution

Answer

The derivative is \( D_x y = -\text{tanh} x \cdot \text{csch}^2 x \).

Step 1: Identify the outer function

The given function is \( y = \ln(\operatorname{coth} x) \). Here, the outer function is \( \ln(u) \), where \( u = \operatorname{coth} x \).

Step 2: Differentiate the outer function

The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \). Therefore, \( \frac{d}{du} \ln(u) = \frac{1}{u} \).

Step 3: Differentiate the inner function

Now differentiate \( u = \operatorname{coth} x \). Recall that \( \operatorname{coth} x = \frac{\cosh x}{\sinh x} \). Its derivative with respect to \( x \) is \( -\text{csch}^2 x \), where \( \text{csch}(x) \) is the hyperbolic cosecant function.

Step 4: Apply the chain rule

According to the chain rule, \( D_{x} y = \frac{dy}{du} \cdot \frac{du}{dx} \). From Steps 2 and 3, we have \( \frac{dy}{du} = \frac{1}{\operatorname{coth} x} \) and \( \frac{du}{dx} = -\text{csch}^2 x \).

Step 5: Simplify the derivative

Now, combine the results from the previous steps: \( D_x y = \frac{1}{\operatorname{coth} x} \cdot (-\text{csch}^2 x) = -\text{csch}^2 x \cdot \frac{\sinh x}{\cosh x} \). Notice that \( \frac{1}{\operatorname{coth} x} = \frac{\sinh x}{\cosh x} = \text{tanh} x \). Thus, \( D_x y = -\text{tanh} x\cdot \text{csch}^2 x \).

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus, especially when dealing with composite functions. It helps in finding the derivative of a function that is nested inside another function. In simple terms, if you have a function inside another function, like in our exercise with \( y = \ln(\operatorname{coth} x) \), the chain rule comes into play.

Here are the main steps involved when using the chain rule:

• Identify the outer and inner functions. In our case, \( \ln(u) \) is the outer function and \( u = \operatorname{coth} x \) is the inner function.
• Differentiate the outer function in terms of the inner function. For \( \ln(u) \), the derivative is \( \frac{1}{u} \).
• Differentiate the inner function concerning the variable, \( x \) here. The derivative of \( \operatorname{coth} x \) is \( -\text{csch}^2 x \).
• Finally, multiply the derivatives you've calculated. According to the chain rule, the derivative \( D_x y \) is \( \frac{dy}{du} \cdot \frac{du}{dx} \).

Applying these steps allows us to tackle derivatives of complex compositions accurately. Understanding each function's role in the chain makes the process straightforward.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for hyperbolas. They are crucial in various areas of calculus, particularly in problems involving exponential growth and decay. The hyperbolic cotangent function, \( \operatorname{coth} x \), is one type, defined as \( \frac{\cosh x}{\sinh x} \), where \( \cosh x \) and \( \sinh x \) are the hyperbolic cosine and sine functions respectively.

These functions include:

• \( \sinh x = \frac{e^x - e^{-x}}{2} \)
• \( \cosh x = \frac{e^x + e^{-x}}{2} \)
• \( \operatorname{coth} x = \frac{\cosh x}{\sinh x} \)
• \( \text{csch}(x) = \frac{1}{\sinh x} \)

These functions have unique derivatives that are quite similar to their trigonometric counterparts. For \( \operatorname{coth} x \), its derivative \( -\text{csch}^2 x \) is akin to the way \( \cot x \) differentiates to \(-\csc^2 x \).

In practice, hyperbolic functions bring ease in solving equations involving exponential and logarithmic components, often appearing in real-world applications like physics and engineering.

Differentiation

Differentiation is the cornerstone of calculus. It involves finding how a function changes as the inputs change, which is determined through the derivative. Practically, this tells us the rate of change or the slope of a function at any given point.

When differentiating, the main goal is to obtain the derivative of a function with respect to its variable. In our example, the aim is to find \( D_x y \), the derivative of \( y = \ln(\operatorname{coth} x) \) with respect to \( x \).

Here's a quick overview of the process:

• First, identify which formulas and rules apply. For composite functions, the chain rule is a go-to method.
• Carry out differentiation on known functions, using their standard derivatives. This includes basic derivatives like \( \frac{d}{dx} x^n = n x^{n-1} \) and advanced ones like \( \frac{d}{dx} \ln(u) = \frac{1}{u} \times u' \), which we used in this exercise.
• Simplify the outcome to make it more comprehensive and often more useful. In our case, recognizing that \( \frac{1}{\operatorname{coth} x} = \text{tanh} x \) helps simplify the solution effectively.

Successful differentiation requires an understanding of functions' behaviors and simple algebraic manipulation, enhancing our ability to model and interpret rates of change.