Problem 18 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\ln (\cosh z)$$

Step-by-Step Solution

Answer

The derivative is \( \tanh(z) \).

Step 1: Identify the Function to Derive

We need to find the derivative of the given function, which is \( y = \ln(\cosh(z)) \). Our goal is to determine \( \frac{dy}{dz} \).

Step 2: Apply the Chain Rule

The function \( y \) is a composition of the natural logarithm and the hyperbolic cosine function. To differentiate it, apply the chain rule: \( \frac{dy}{dz} = \frac{d}{dz}[\ln(\cosh(z))] = \frac{1}{\cosh(z)} \cdot \frac{d}{dz}[\cosh(z)] \).

Step 3: Differentiate the Inner Function

Now, differentiate \( \cosh(z) \). The derivative with respect to \( z \) is \( \sinh(z) \) because the derivative of \( \cosh(z) \) is \( \sinh(z) \). Therefore, \( \frac{d}{dz}[\cosh(z)] = \sinh(z) \).

Step 4: Combine Results

Substituting the derivative of the inner function back into our expression from Step 2, we have \( \frac{dy}{dz} = \frac{1}{\cosh(z)} \cdot \sinh(z) \).

Step 5: Simplify the Expression

Recall the identity \( \tanh(z) = \frac{\sinh(z)}{\cosh(z)} \). Using this identity, simplify the derivative to \( \frac{dy}{dz} = \tanh(z) \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. Composite functions are functions nested within each other, like the function in our exercise, where we have a natural logarithm (ln) applied to a hyperbolic cosine function (cosh). The Chain Rule provides a method to tackle such problems by breaking them down into simpler parts.

• When applying the Chain Rule, you first identify the outer and inner functions. In our example: outer function is \( \ln(x)\) and inner function is \( \cosh(z)\).
• Differentiate the outer function with respect to the inner function, then multiply by the derivative of the inner function with respect to the variable.

For our exercise: Differentiate \( \ln(u) \, u = \cosh(z)\) yielding \( \frac{1}{u}\frac{du}{dz}\), and since \( u = \cosh(z)\), you then multiply by \( \frac{d}{dz} [\cosh(z)] = \sinh(z)\). This process gives the complete derivative.

Hyperbolic Functions

Hyperbolic functions share similarities with trigonometric functions and are especially useful in calculus for handling exponential expressions. In the earlier exercise, we encountered the hyperbolic cosine \( \cosh(z)\). Understanding its properties and derivatives is crucial when working with calculus problems involving these functions.

• The hyperbolic cosine, \( \cosh(z) = \frac{e^z + e^{-z}}{2}\), is one of the basic hyperbolic functions.
• The derivative of \( \cosh(z)\) is \( \sinh(z)\) (hyperbolic sine), which is another foundational hyperbolic function, defined as \( \sinh(z) = \frac{e^z - e^{-z}}{2}\).
• An essential identity to remember is \( \tanh(z) = \frac{\sinh(z)}{\cosh(z)}\).

These identities and properties enable simplification and computation of derivatives and integrals involving hyperbolic functions.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate certain types of functions by utilizing the natural logarithm's properties. This technique can simplify the differentiation of products, quotients, or powers that might otherwise be complex.

• In the context of our example function \( y=\ln(\cosh(z))\), you apply the natural logarithm directly to the function.
• When differentiating \( \ln(\cosh(z))\), differentiate the logarithm part and use the Chain Rule for the implicit \( \cosh(z)\).
• This results in applying \( \frac{1}{u} \frac{du}{dz}\) where \( u=\cosh(z)\), which allows us to express the differentiation conveniently.

Logarithmic differentiation is especially advantageous when dividing complex algebraic expressions involving exponents or products of several functions. It often makes calculations more straightforward and less error-prone.