Question

Find the derivatives of the given functions and graph along with the function to ensure your answer is correct.\([T] \ln (\operatorname{sech}(x)+\tanh (x))\)

Short answer

The derivative is \( \frac{-\operatorname{sech}(x)\tanh(x) + \operatorname{sech}^2(x)}{\operatorname{sech}(x) + \tanh(x)} \).

Step-by-step solution

Identify the Function Composition

The given function is \( f(x) = \ln(\operatorname{sech}(x) + \tanh(x)) \). Note that this is a composition of functions involving logarithms and hyperbolic trigonometric functions. We need to apply the chain rule to differentiate this.

Differentiate the Outer Function

The outer function is \( \ln(u) \) where \( u = \operatorname{sech}(x) + \tanh(x) \). The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \).

Differentiate the Inner Function

The inner function is \( u(x) = \operatorname{sech}(x) + \tanh(x) \). Differentiate each component: \( \operatorname{sech}(x) = \frac{1}{\cosh(x)} \) and \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \).- The derivative of \( \operatorname{sech}(x) \) is \( -\operatorname{sech}(x)\tanh(x) \).- The derivative of \( \tanh(x) \) is \( \operatorname{sech}^2(x) \). Thus, \( u'(x) = -\operatorname{sech}(x)\tanh(x) + \operatorname{sech}^2(x) \).

Apply the Chain Rule

Using the chain rule, \( f'(x) = \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot u'(x) \). Substitute the expressions from Steps 2 and 3:\[ f'(x) = \frac{-\operatorname{sech}(x)\tanh(x) + \operatorname{sech}^2(x)}{\operatorname{sech}(x) + \tanh(x)} \].

Simplify the Derivative

To simplify, consider common terms in the numerator and express the function in terms involving \( \cosh(x) \) and \( \sinh(x) \) if needed. This can help verify algebraic steps, though not required.

Graph the Functions

Graph the original function \( f(x) = \ln(\operatorname{sech}(x) + \tanh(x)) \) and its derivative \( f'(x) = \frac{-\operatorname{sech}(x)\tanh(x) + \operatorname{sech}^2(x)}{\operatorname{sech}(x) + \tanh(x)} \) on a coordinate plane. Ensure that the graphs reflect the critical points and zero-crossings of the derivative.

Key concepts

Chain Rule

When dealing with derivatives, one of the most powerful tools in calculus is the chain rule. This rule is incredibly useful when you have a function composed of other functions. In simpler terms, when you have a function within a function.

Think of the chain rule as a way to "zoom in" on the different layers of functions and differentiate each one step by step. The general formula for the chain rule is:

• If you have a composed function given by \( f(g(x)) \), then the derivative \( f'(x) = f'(g(x)) \cdot g'(x) \).

In this format, \( f(g(x)) \) means that \( f \) is your outer function and \( g(x) \) is your inner function.

For our problem, the outer function is the natural logarithm, \( \,\ln \,\), and the inner function is the combination of hyperbolic functions. Understanding how to apply the chain rule lets you break down the differentiation process into manageable steps.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but are based on hyperbolas rather than circles. They have similar properties to the familiar sine and cosine functions, but they use hyperbolic sine \( \sinh(x) \) and hyperbolic cosine \( \cosh(x) \).

In calculus, hyperbolic functions often appear in problems involving exponential functions. For instance:

• \( \,\operatorname{sech}(x) \) is the hyperbolic secant, calculated as \( \frac{1}{\cosh(x)} \).
• \( \,\tanh(x) \) is the hyperbolic tangent, calculated as \( \frac{\sinh(x)}{\cosh(x)} \).

These functions behave remarkably like their trigonometric cousins. They have derivatives that are important in calculus.

• The derivative of \( \operatorname{sech}(x) \) is \( -\operatorname{sech}(x)\tanh(x) \).
• The derivative of \( \tanh(x) \) is \( \operatorname{sech}^2(x) \).

These derivatives are fundamental for deriving more complex functions when hyperbolic functions are involved.

Logarithmic Differentiation

Logarithmic differentiation is a technique often used when differentiating functions that are products, quotients, or have an exponent. Utilizing the properties of logarithms can simplify the differentiation process.

In mathematical terms, if you have a function \( f(x) \) that can be expressed as \( \,\ln(f(x)) \,\), using logarithmic differentiation allows you to differentiate the log of the function instead of the function itself.

• For \( \ln(u(x)) \), the differentiation is straightforward: \( \frac{1}{u(x)} \cdot u'(x) \).

In the context of our problem, logarithmic differentiation allows us to handle the outer layer of our composite function easily. It provides a nice simplification by letting us focus on the inner composite functions and their derivatives first.

Function Composition

Function composition is when you combine two or more functions into a single, compounded function. This is expressed as \( f(g(x)) \), where \( g(x) \) is applied first, followed by \( f \).

In problems like the one we've looked at, understanding how to separate out and manage each component function is crucial. Composition often arises in calculus when dealing with more complex functions built from simpler ones. Here’s how to approach such problems:

• Identify the outer and inner functions. The outer function here is \( \ln \), and the inner one is \( \operatorname{sech}(x) + \tanh(x) \).
• Apply the chain rule for differentiation, as this helps break down the complex derivative into simpler parts.

Appreciating function composition not only helps in differentiation but also deepens your understanding of how different mathematical operations can be interconnected.