Problem 75 - Single Variable Calculus

Question

Prove that \(\frac{d}{d x} \operatorname{sech} u=-(\operatorname{sech} u \tanh u) \frac{d u}{d x}\).

Step-by-Step Solution

Answer

To prove that $\frac{d}{d x} \operatorname{sech} u=-(\operatorname{sech} u \tanh u) \frac{d u}{d x}$, we apply the chain rule to find the derivative of $\operatorname{sech}u$ with respect to $x$, calculate the derivative of $\frac{1}{\cosh u}$ with respect to $u$, and express the result in terms of sech and tanh. Following these steps, we obtain the derivative as $- (\operatorname{sech} u \tanh u) \frac{du}{dx}$.

Step 1: Recall the definition of sech and tanh functions

the given exercise involves the sech and tanh functions, which are defined as: \[\operatorname{sech} u = \frac{1}{\cosh u}\] \[\tanh u = \frac{\sinh u}{\cosh u}\]

Step 2: Use the chain rule

In order to calculate the derivative of the given function \(\operatorname{sech} u\) with respect to \(x\), we use the chain rule. The chain rule states that if we have a composite function \(y(u(x))\), the derivative of \(y\) with respect to \(x\) is: \[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\] Applying the chain rule to our function, we get: \[\frac{d(\operatorname{sech} u)}{dx} = \frac{d(\operatorname{sech} u)}{du} \cdot \frac{du}{dx}\]

Step 3: Calculate the derivative of sech with respect to u

Now, we need to compute the derivative of the function \(\operatorname{sech} u\) with respect to the variable \(u\). We can rewrite \(\operatorname{sech} u\) as: \[\operatorname{sech} u = \frac{1}{\cosh u}\] Now, using the basic rules of differentiation, the derivative of the function \(\frac{1}{\cosh u}\) with respect to \(u\) is: \[\frac{d}{du}\left(\frac{1}{\cosh u}\right) = -\frac{1}{\cosh^2 u} \frac{d(\cosh u)}{du}\] Recall that the derivative of \(\cosh u\) is \(\sinh u\), so we can simplify it to: \[\frac{d(\operatorname{sech} u)}{du} = -\frac{\sinh u}{\cosh^2 u}\]

Step 4: Express the derivative in terms of sech and tanh

Since we know the definitions of the sech and tanh functions, we can rewrite the expression for the derivative in terms of these functions: \[-\frac{\sinh u}{\cosh^2 u} = -(\operatorname{sech} u \tanh u)\]

Step 5: Combine the results

Finally, we can combine the results of Steps 2, 3, and 4 to write the derivative of the function \(\operatorname{sech} u\) with respect to \(x\): \[\frac{d(\operatorname{sech} u)}{dx} = \left(-(\operatorname{sech} u \tanh u)\right) \frac{du}{dx}\] This proves the given statement.

Key Concepts

Chain Rule

The chain rule is a fundamental theorem in calculus used for finding the derivative of composite functions. It is essential when dealing with functions that are nested or interlinked, such as when a function exists inside another function. Think of it like peeling an onion: you handle one layer at a time.

For functions given as \( y(u(x)) \), the chain rule states:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

This expression indicates that to find the derivative of \(y\) with respect to \(x\), we first differentiate \(y\) with respect to \(u\), and then multiply it by the derivative of \(u\) with respect to \(x\).

In practical terms, when looking at expressions like \( \operatorname{sech} u \), it’s crucial to apply the chain rule to ensure the differentiation is accurate when functions are dependent on other variables.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola rather than a circle. These functions arise frequently in calculus, physics, and engineering.

Some basic hyperbolic functions include:

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

The hyperbolic secant function, expressed as \( \operatorname{sech} u = \frac{1}{\cosh u} \), mimics the trigonometric secant but is adapted for hyperbolic contexts.

These hyperbolic functions satisfy similar identities to their circular counterparts, such as \( \cosh^2 u - \sinh^2 u = 1 \), which is reminiscent of the Pythagorean identity in trigonometry.

Sech Function Differentiation

Differentiating the \( \operatorname{sech} \) function may seem complex at first glance but becomes straightforward with a step-by-step approach. The function is defined as:

• \( \operatorname{sech} u = \frac{1}{\cosh u} \)

To differentiate \( \operatorname{sech} u \) with respect to \( u \), use the quotient rule—since it is the reciprocal of a hyperbolic function:

• \( \frac{d}{du} \left( \frac{1}{\cosh u} \right) = - \frac{\sinh u}{\cosh^2 u} \)

By referring back to the definition of \( \tanh u \):

• \( \tanh u = \frac{\sinh u}{\cosh u} \)

You can confirm that:

• \( - \frac{\sinh u}{\cosh^2 u} = -\operatorname{sech} u \tanh u \)

Thus, proving the differentiation involves combining basic differential rules with hyperbolic identities.

Calculus Proof

A calculus proof involves logical reasoning to establish the truth of a mathematical statement. In this context, we proved the derivative of \( \operatorname{sech} u \) by methodically following established calculus principles.

The exercise required:

• Understanding the \( \operatorname{sech} \) function from its definition, \( \operatorname{sech} u = \frac{1}{\cosh u} \).
• Applying differentiation rules like the chain and quotient rules.
• Relating the derivative back to other hyperbolic functions \( \tanh u = \frac{\sinh u}{\cosh u} \).

By substituting back into the main derivative formula, we completed the proof by showing it equals \( -(\operatorname{sech} u \tanh u) \cdot \frac{du}{dx} \). This systematic approach ensures every step logically leads to the final proved result.