Problem 26 - Essential Calculus Early Transcendentals

Question

Find the derivative of the function. Simplify where possible. $$ f(x)=\tanh \left(1+e^{2 x}\right) $$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = 2e^{2x} \cdot \text{sech}^2(1 + e^{2x}) \).

Step 1: Understand the Function

We are given the function \( f(x) = \tanh(1 + e^{2x}) \). Our goal is to find the derivative, \( f'(x) \). The function is composed of the hyperbolic tangent function applied to \( 1 + e^{2x} \).

Step 2: Use Chain Rule

The derivative of a composite function \( g(h(x)) \) is \( g'(h(x)) \cdot h'(x) \). Here, \( g(u) = \tanh(u) \) and \( h(x) = 1 + e^{2x} \). We will differentiate each part separately and then apply the chain rule.

Step 3: Differentiate \( \tanh(u) \)

The derivative of \( \tanh(u) \) with respect to \( u \) is \( \text{sech}^2(u) \). So, \( g'(u) = \text{sech}^2(u) \).

Step 4: Differentiate \( 1 + e^{2x} \)

The derivative of \( 1 + e^{2x} \) with respect to \( x \) is \( 2e^{2x} \) because the derivative of \( e^{2x} \) is \( 2e^{2x} \) and the constant 1 has a derivative of 0.

Step 5: Apply the Chain Rule

Combine the derivatives using the chain rule: \[ f'(x) = g'(h(x)) \cdot h'(x) = \text{sech}^2(1 + e^{2x}) \cdot 2e^{2x}. \]

Step 6: Simplify the Expression

The expression for \( f'(x) \) is \[ f'(x) = 2e^{2x} \cdot \text{sech}^2(1 + e^{2x}). \] There are no further simplifications possible, so this is our final answer.

Key Concepts

Hyperbolic Functions

Hyperbolic functions are analogs of the trigonometric functions but for a hyperbola, instead of a circle. The hyperbolic tangent function, denoted as \( \tanh(x) \), is one of these functions. It's defined as the ratio of the hyperbolic sine and hyperbolic cosine: \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \).
These functions often appear in calculus, providing a useful tool for modeling certain behaviors in mathematics and physics.
Some common hyperbolic functions include:

• Hyperbolic sine: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• Hyperbolic cosine: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
• Hyperbolic tangent: \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \)

For our function, \( f(x) = \tanh(1 + e^{2x}) \), we apply the properties of hyperbolic functions during differentiation. Particularly, we use the fact that the derivative of \( \tanh(u) \) is \( \text{sech}^2(u) \), where \( \text{sech}(x) = \frac{1}{\cosh(x)} \).
This highlights how hyperbolic functions and their derivatives can be utilized to solve calculus problems efficiently.

Chain Rule

The chain rule is a fundamental differentiation technique used for finding the derivative of composite functions. A composite function is essentially a "function within a function," and its structure often necessitates a direct method to differentiate.
For a function \( g(h(x)) \), the chain rule states that its derivative is \( g'(h(x)) \cdot h'(x) \). This rule helps us break down the process of differentiation into manageable parts, especially when dealing with complex functions.

Application of the Chain Rule

In our exercise, we have the composite function \( \tanh(1 + e^{2x}) \). Here:

• \( g(u) = \tanh(u) \) with derivative \( g'(u) = \text{sech}^2(u) \)
• \( h(x) = 1 + e^{2x} \) with derivative \( h'(x) = 2e^{2x} \)

By applying the chain rule, we compute the derivative with respect to \( x \) as \( f'(x) = \text{sech}^2(1 + e^{2x}) \cdot 2e^{2x} \).
The chain rule simplifies the process by allowing us to handle each component separately and then combine them.

Differentiation Techniques

Differentiation techniques are essential tools for calculating derivatives and understanding the rate of change of a function. These methods range from basic power rules to more advanced applications involving trigonometric, exponential, and hyperbolic functions.
Some of the key differentiation techniques include:

• Power Rule: Used when differentiating functions of the form \( x^n \). The derivative is \( nx^{n-1} \).
• Product Rule: Useful for differentiating products of two functions.
• Quotient Rule: Applied when dealing with quotients of two functions.
• Chain Rule: Vital for composite functions, as explained earlier.

In our problem, besides applying the chain rule, we also differentiated an exponential function. The derivative of \( e^{2x} \) is \( 2e^{2x} \), illustrating the rule that \( \frac{d}{dx}(e^{ax}) = ae^{ax} \) for any constant \( a \).
Understanding these techniques is crucial for effectively solving complex calculus problems that involve combinations of different types of functions.