Problem 20 - Calculus Early Transcendentals

Question

Derive the following derivative formulas given that \(d / d x(\cosh x)=\sinh x\) and \(d / d x(\sinh x)=\cosh x\). \(d / d x(\operatorname{sech} x)=-\operatorname{sech} x \tanh x\)

Step-by-Step Solution

Answer

The derivative of the secant hyperbolic function \(\operatorname{sech}(x)\) is \(-\operatorname{sech}(x) \tanh(x)\).

Step 1: Recall the definition of \(\operatorname{sech}(x)\)

Recall that the secant hyperbolic function, or \(\operatorname{sech}(x)\), is defined as the reciprocal of the hyperbolic cosine function: $$ \operatorname{sech}(x) = \frac{1}{\cosh(x)} $$

Step 2: Use the chain rule to differentiate \(\operatorname{sech}(x)\)

To find the derivative of \(\operatorname{sech}(x)\) with respect to \(x\), we can use the chain rule. The chain rule states that if we have a function \(y(u)\) and \(u(x)\), then the derivative of \(y(u(x))\) with respect to \(x\) is: $$ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} $$ In our case, \(y(u) = \frac{1}{u}\) and \(u(x) = \cosh(x)\). Thus, we need to find \(\frac{dy}{du}\) and \(\frac{du}{dx}\).

Step 3: Differentiate \(y(u)\) and \(u(x)\)

To find \(\frac{dy}{du}\) and \(\frac{du}{dx}\), we need to differentiate the functions \(y(u) = \frac{1}{u}\) and \(u(x) = \cosh(x)\). Recall that \(\frac{d}{dx}\cosh x = \sinh x\), so we have: $$ \frac{du}{dx} = \frac{d}{dx}(\cosh(x)) = \sinh(x) $$ Now, differentiate \(y(u) = \frac{1}{u}\): $$ \frac{dy}{du} = -\frac{1}{u^2} $$

Step 4: Apply the chain rule and simplify

Now we can apply the chain rule from step 2: $$ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = -\frac{1}{u^2} \cdot \sinh(x) $$ Replace \(u\) with the function \(\cosh(x)\): $$ \frac{dy}{dx} = -\frac{1}{(\cosh(x))^2} \cdot \sinh(x) $$ Recall that \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\), so \((\operatorname{sech}(x))^2 = \frac{1}{(\cosh(x))^2}\). We can now rewrite the derivative as: $$ \frac{dy}{dx} = -(\operatorname{sech}(x))^2 \cdot \sinh(x) $$

Step 5: Find the \(\tanh(x)\) expression

Recall that the hyperbolic tangent function is defined as \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\). We can rewrite this as: $$ \sinh(x) = \tanh(x) \cdot \cosh(x) $$

Step 6: Substitute the \(\sinh(x)\) expression into the derivative

Now, substitute the expression for \(\sinh(x)\) from step 5 into the derivative: $$ \frac{dy}{dx} = -(\operatorname{sech}(x))^2 \cdot \tanh(x) \cdot \cosh(x) $$ Recall that \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\), so we can simplify the derivative: $$ \frac{dy}{dx} = -\operatorname{sech}(x) \cdot \tanh(x) $$ So, the derivative of \(\operatorname{sech}(x)\) is: $$ \frac{d}{dx}(\operatorname{sech}(x)) = -\operatorname{sech}(x) \tanh(x) $$

Key Concepts

Chain Rule

The chain rule is a fundamental principle in calculus, essential for finding derivatives of composite functions. When you have a function that is built from two or more functions, you cannot simply find the derivative by differentiating each part individually. Instead, you need to use the chain rule. This rule tells you how to handle these composite situations.

Here's how it works: if you have a function composed as \( y = f(g(x)) \), the derivative of \( y \) with respect to \( x \) is found by multiplying the derivative of \( f \) with respect to \( g \), noted as \( \frac{dy}{dg} \), by the derivative of \( g \) with respect to \( x \), noted as \( \frac{dg}{dx} \). Thus,\[ \frac{dy}{dx} = \frac{dy}{dg} \cdot \frac{dg}{dx} \]

In the context of hyperbolic functions, like \( \operatorname{sech}(x) \), the chain rule is crucial because the derivative involves differentiating \( y(u) = \frac{1}{u} \) and replacing \( u \) with \( \cosh x \). Understanding and applying the chain rule enables you to simplify the computation of complex derivatives efficiently.

Hyperbolic Functions

Hyperbolic functions are analogs of the familiar trigonometric functions but are based on hyperbolas rather than circles. These functions include \( \sinh(x) \) and \( \cosh(x) \), which can be thought of as counterparts to sine and cosine functions.

These functions are important in various areas of calculus, especially when dealing with differential equations. For example:

• \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

Each hyperbolic function has a derivative that is reminiscent of its trigonometric counterpart:

• \( \frac{d}{dx}( \sinh(x) ) = \cosh(x) \)
• \( \frac{d}{dx}( \cosh(x) ) = \sinh(x) \)

Understanding these derivatives is key to solving problems involving hyperbolic functions in calculus, like finding the derivative of \( \operatorname{sech}(x) \). The relation \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \) is also useful because it relates these functions in a way that simplifies calculations involving derivatives.

Differentiation Rules

Differentiation rules are the basic guidelines used to perform differentiation in calculus. They include rules for sums, products, quotients, and chain operations. These expansive rules simplify the process of finding derivatives for complex functions.

Key rules include:

• The Power Rule: \( \frac{d}{dx}(x^n) = n \cdot x^{n-1} \)
• The Product Rule: If \( u(x) \) and \( v(x) \) are differentiable, \( \frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v' \)
• The Quotient Rule: For \( \frac{u}{v} \), \( \frac{d}{dx}(\frac{u}{v}) = \frac{u' \cdot v - u \cdot v'}{v^2} \)

In the context of differentiating \( \operatorname{sech}(x) \), which is \( \frac{1}{\cosh(x)} \), you'd apply the quotient rule due to its reciprocal form. Combined with the chain rule and knowledge of derivatives of \( \sinh(x) \) and \( \cosh(x) \), you can solve complex derivative problems more easily.

These rules form the backbone of derivative computation, allowing for the exploration of various functions' behaviors in calculus.