Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\tanh (\cot x) $$

Step-by-Step Solution

Answer

\(-\text{sech}^2(\cot x) \cdot \csc^2 x\) is the derivative of \(y = \tanh(\cot x)\).

Step 1: Understand the Function

The function we are given is a composition of two functions, \(y = \tanh (\cot x)\). This means we have to apply the chain rule for derivatives. First, identify the inner function \(u = \cot x\) and the outer function \(y = \tanh u\).

Step 2: Differentiate the Outer Function

Differentiate the tangent hyperbolic function with respect to its argument \(u\). Recall that the derivative of \(\tanh u\) is \(\text{sech}^2(u)\), where \(\text{sech}^2(u) = 1/\cosh^2(u)\). So, \(\frac{dy}{du} = \text{sech}^2(u)\).

Step 3: Differentiate the Inner Function

The inner function is \(u = \cot x\). The derivative of \(\cot x\) is \(-\csc^2 x\). Thus, \(\frac{du}{dx} = -\csc^2 x\).

Step 4: Apply the Chain Rule

Use the chain rule to find \(\frac{dy}{dx}\) by multiplying \(\frac{dy}{du}\) and \(\frac{du}{dx}\).\[ D_x y = \frac{dy}{du} \cdot \frac{du}{dx} = \text{sech}^2(\cot x) \cdot (-\csc^2 x) \].

Step 5: Simplify the Expression

Combine the results from Step 4. The derivative of \(y = \tanh(\cot x)\) becomes:\[ \frac{dy}{dx} = -\text{sech}^2(\cot x) \cdot \csc^2 x \].

Key Concepts

Chain Rule

In calculus, the chain rule is a technique used to differentiate composite functions. It plays a crucial role when you have a function inside another function, much like the layers of an onion. To use the chain rule, you identify the inner and outer functions:

• The inner function is the one nested inside another. In this problem, it's the \( \cot x\ \) function.
• The outer function operates on the result of the inner function. Here, it is \( \tanh u \), where \( u = \cot x \).

To apply the chain rule, follow these steps:

• First, differentiate the outer function with respect to the inner function, treating whatever's inside it as a variable.
• Then, multiply this result by the derivative of the inner function with respect to \(x\).

This "two-step" process allows you to handle complex derivatives by breaking them down into manageable pieces.
Thus, the chain rule formula, simplified, is: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \] where each derivative is found independently and then multiplied together.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for hyperbolas instead of circles. They are defined using exponential functions and are significant in many mathematical and physical applications. In this exercise, the specific hyperbolic function is the hyperbolic tangent, \( \tanh x \).
A few key hyperbolic functions include:

• Tanh (\( \tanh x \)): Defined as \( \tanh x = \frac{\sinh x}{\cosh x} \), where \( \sinh x = \frac{e^x - e^{-x}}{2} \) (sinh) and \( \cosh x = \frac{e^x + e^{-x}}{2} \) (cosh).
• Sech (\( \text{sech} \,x \)): It is defined as \( \text{sech} \, x = \frac{1}{\cosh x} \).

The derivative of the hyperbolic tangent is particularly important: \( \text{sech}^2(x) \).This derivative shows how \( \tanh \) changes with its input and is derived from its definition as a quotient of \( \sinh \) and \( \cosh \).When working with hyperbolic functions, remember they behave similar to trigonometric functions, but their properties derive from hyperbolas rather than circles.

Trigonometric Functions

Trigonometric functions relate angles of triangles to the lengths of their sides; they are fundamental in geometry, physics, and engineering. The function \( \cot x \) is involved in this exercise as the inner function in our chain rule application.
Key trigonometric functions include:

• Sine (\( \sin \)): Relates to the opposite side over the hypotenuse in a right triangle.
• Cosine (\( \cos \)): Defines the adjacent side over the hypotenuse.
• Cotangent (\( \cot \)): The reciprocal of tangent, so \( \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x} \).
• Cosecant (\( \csc \)): The reciprocal of sine, \( \csc x = \frac{1}{\sin x} \).

In this problem's context, the derivative of \( \cot x \) is \( -\csc^2 x \). This result is crucial for applying the chain rule and illustrates how trigonometric derivatives often involve reciprocal functions. Understanding these relationships helps dissect complex calculus problems.