Problem 43 - Essential Calculus Early Transcendentals

Question

Show that \(\frac{d}{d x} \arctan (\tanh x)=\operatorname{sech} 2 x\)

Step-by-Step Solution

Answer

Yes, the derivative is \(\operatorname{sech} 2x\).

Step 1: Differentiate the Function

To find \(\frac{d}{dx} \arctan (\tanh x)\), use the chain rule. The derivative of \(\arctan u\) with respect to \(u\) is \(\frac{1}{1+u^2}\), and by the chain rule, \(\frac{d}{dx} \arctan (\tanh x) = \frac{1}{1+\tanh^2 x} \cdot \frac{d}{dx} (\tanh x)\).

Step 2: Derivative of Hyperbolic Tangent

The derivative of \(\tanh x\) is \(\operatorname{sech}^2 x\). Therefore, substituting back, the derivative becomes \(\frac{\operatorname{sech}^2 x}{1+\tanh^2 x}\).

Step 3: Simplify the Expression

Recall the identity \(\operatorname{sech}^2 x = 1 - \tanh^2 x\). This implies that \(\operatorname{sech}^2 x = \frac{1}{\cosh^2 x}\). To rewrite the denominator, use \(1 + \tanh^2 x = \frac{1}{\cosh^2 x}\). Thus, \(\frac{\operatorname{sech}^2 x}{1 + \tanh^2 x} = \frac{1}{\cosh^2 x}\).

Step 4: Verify with Given Function

Given \(\operatorname{sech} 2x\), write it as \(\operatorname{sech}(2x) = \frac{2}{e^{2x} + e^{-2x}}\). Using hyperbolic identities, we know \(\cosh(2x) = \cosh^2 x + \sinh^2 x\). Therefore, \(\operatorname{sech}^2 x = \operatorname{sech}(2x)\). This matches our simplified form, confirming the derivatives are equivalent.

Key Concepts

Hyperbolic Functions

Hyperbolic functions, much like their trigonometric counterparts, have many useful properties in calculus. They are defined using exponential functions and include functions such as hyperbolic sine

• \(\sinh x = \frac{e^x - e^{-x}}{2}\)
• hyperbolic cosine \(\cosh x = \frac{e^x + e^{-x}}{2}\)
• hyperbolic tangent \(\tanh x = \frac{\sinh x}{\cosh x}\)
• and hyperbolic secant \(\operatorname{sech} x = \frac{1}{\cosh x}\)

Hyperbolic functions act much like trigonometric functions and are crucial in solving many calculus problems.

One key identity we use often is \(\operatorname{sech}^2 x = 1 - \tanh^2 x\), reminiscent of \(\cos^2 x + \sin^2 x = 1\) for circular functions.

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivatives of composite functions. It states that if you have a function \(y = f(g(x))\), the derivative \(\frac{dy}{dx}\) can be found by multiplying the derivative of the outer function by the derivative of the inner function: \[\frac{dy}{dx} = f'(g(x)) \cdot g'(x).\]Using the chain rule allows us to find derivatives of complex nested functions efficiently.

In our exercise, \(\arctan(\tanh(x))\) requires the chain rule because it involves two functions: the arctan as the outer function and tanh as the inner. We differentiate the arctan function first, then multiply it by the derivative of the tanh function.

This step-by-step method ensures that nothing gets overlooked in differentiation and helps simplify each part separately.

Calculus Problem Solving

Solving calculus problems involves a strategic approach where understanding and applying fundamental concepts are key.

To differentiate \(\arctan(\tanh(x))\), we:

• Made use of the chain rule to manage composite functions.
• Found the necessary derivatives, particularly that \(\operatorname{sech}^2 x\) is a simple transformation of \(\tanh x\).
• Employed hyperbolic identities to simplify expressions efficiently.

A clear understanding of fundamental topics such as derivatives, the chain rule, and function identities provides a straightforward path to solving these exercises.

Whenever facing complex calculus problems, breaking them into smaller manageable parts and systematically applying these principles leads to the solution.