Problem 21 - Thomas Calculus

Question

In Exercises \(13-24,\) find the derivative of \(y\) with respect to the appropriate variable. $$ y=\ln \cosh v-\frac{1}{2} \tanh ^{2} v $$

Step-by-Step Solution

Answer

The derivative is \( \tanh^3 v \).

Step 1: Differentiate the first term

The first term is \( \ln(\cosh v) \). We will apply the chain rule here. The derivative of \( \ln(u) \) with respect to \( u \) is \( \frac{1}{u} \), and the derivative of \( \cosh(v) \) with respect to \( v \) is \( \sinh(v) \). Therefore, using the chain rule, the derivative of \( \ln(\cosh v) \) with respect to \( v \) is \( \frac{\sinh v}{\cosh v} = \tanh v \).

Step 2: Differentiate the second term

The second term is \( -\frac{1}{2} \tanh^2 v \). Use the chain rule here again. The derivative of \( \tanh v \) with respect to \( v \) is \( \text{sech}^2 v \). So, the derivative of \( \tanh^2 v \) is \( 2 \tanh v \cdot \text{sech}^2 v \). Now, multiplying by the constant \(-\frac{1}{2}\), the derivative becomes \( -\tanh v \cdot \text{sech}^2 v \).

Step 3: Combine the derivatives

Add the derivatives from Step 1 and Step 2 to find the derivative of the entire expression. The derivative of \( y \) is given by \( \tanh v - \tanh v \cdot \text{sech}^2 v \).

Step 4: Simplify the expression

Notice that the expression \( \tanh v - \tanh v \cdot \text{sech}^2 v \) can be simplified. Since \( \text{sech}^2 v = 1 - \tanh^2 v \), the expression simplifies to \( \tanh v \left( 1 - (1 - \tanh^2 v) \right) = \tanh^3 v \). Thus, the simplified derivative is \( \tanh^3 v \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find the derivative of a composite function. Imagine you have a function inside another function. The chain rule helps you differentiate it by applying the rule step by step.
To use the chain rule:

• Find the derivative of the outer function, treating the inner function as a single variable.
• Multiply by the derivative of the inner function.

In our example, to differentiate \( \ln(\cosh v) \), we see it as a combination of \( \ln(u) \) and \( u = \cosh v \). First, derive \( \ln(u) \) with respect to \( u \), which is \( \frac{1}{u} \).
Next, derive \( \cosh v \) with respect to \( v \), giving \( \sinh v \).
Combine these by multiplying the two derivatives: \( \frac{\sinh v}{\cosh v} = \tanh v \). This step leads us to the derivative of the first term of our function.

Hyperbolic Functions

Hyperbolic functions, like \( \cosh \) and \( \tanh \), are analogs of trigonometric functions but for hyperbolas. They often appear in calculus because of their smooth properties and how they relate to each other.
For example:

• The derivative of \( \cosh v \) is \( \sinh v \).
• \( \tanh v \) is the hyperbolic tangent function equivalent to \( \frac{\sinh v}{\cosh v} \).
• The derivative of \( \tanh v \) is \( \text{sech}^2 v \) where \( \text{sech} \, v = \frac{1}{\cosh v} \).

In the original problem, recognizing these derivatives was key to solving the exercise.
Hyperbolic functions have properties that often simplify problems, similar to how sine and cosine reduce trigonometric equations.

Derivative Simplification

Simplifying a derivative often makes it more understandable and easier to work with.
In our example, the derivative \( \tanh v - \tanh v \cdot \text{sech}^2 v \) may seem complex at first. But knowing that \( \text{sech}^2 v = 1 - \tanh^2 v \) allows us to simplify the expression.
Here's the breakdown:

• Start by understanding that simplification can make checking your work easier and the results more intuitive.
• Use known identities, like \( \text{sech}^2 v = 1 - \tanh^2 v \).
• This simplification converts \( \tanh v - \tanh v \cdot \text{sech}^2 v \) into \( \tanh v \left( 1 - (1 - \tanh^2 v) \right) = \tanh^3 v \).

The final result, \( \tanh^3 v \), neatly showcases the benefit of simplification in calculus, making derivatives more manageable and understandable.