Problem 19 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\ln \left(\frac{e^{\theta}}{1+e^{\theta}}\right)$$

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{1 + e^{\theta}} \).

Step 1: Identify the Rule to Use

The function is given as \( y = \ln\left(\frac{e^{\theta}}{1+e^{\theta}}\right) \). The derivative of a natural logarithm function is \( \frac{1}{u} \cdot \frac{du}{d\theta} \), where \( u \) is the argument of the log. We will apply the chain rule and the quotient rule to find \( \frac{du}{d\theta} \).

Step 2: Define and Differentiate the Inner Function

Let \( u = \frac{e^{\theta}}{1+e^{\theta}} \). We will differentiate \( u \) using the quotient rule. The quotient rule states \( \left(\frac{v}{w}\right)' = \frac{v'w - vw'}{w^2} \). Here \( v = e^{\theta} \) and \( w = 1 + e^{\theta} \).

Step 3: Apply the Quotient Rule

First, find the derivatives \( v' = e^{\theta} \) and \( w' = e^{\theta} \). Apply the quotient rule: \( \frac{du}{d\theta} = \frac{e^{\theta}(1 + e^{\theta}) - e^{\theta} \, e^{\theta}}{(1 + e^{\theta})^2} = \frac{e^{\theta}}{(1 + e^{\theta})^2} \).

Step 4: Differentiate the Logarithmic Function

The derivative of \( y \) is then \( \frac{1}{u} \cdot \frac{du}{d\theta} \). Substitute \( u = \frac{e^{\theta}}{1 + e^{\theta}} \) and \( \frac{du}{d\theta} = \frac{e^{\theta}}{(1 + e^{\theta})^2} \) to get: \[ \frac{1}{\frac{e^{\theta}}{1 + e^{\theta}}} \cdot \frac{e^{\theta}}{(1 + e^{\theta})^2} = \frac{1 + e^{\theta}}{e^{\theta}} \cdot \frac{e^{\theta}}{(1 + e^{\theta})^2} = \frac{1}{1 + e^{\theta}}. \]

Step 5: Simplify the Expression

The expression \( \frac{1}{1 + e^{\theta}} \) is already simplified. This is the derivative of the given function with respect to \( \theta \).

Key Concepts

Natural Logarithm Function

The natural logarithm, denoted as \( \ln(x) \), is an important function in calculus and appears frequently in derivatives. It expresses the power to which \( e \) (approximately 2.718) must be raised to produce a given number. In differentiation, the natural logarithm has a very neat rule for its derivative: for a function \( y = \ln(u) \), its derivative with respect to \( x \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).

It's crucial to recognize that when the argument of a natural logarithm is itself a function \( u(x) \), we need to use additional rules such as the chain rule to find the overall derivative. In our exercise, the argument is a fraction, \( \frac{e^{\theta}}{1+e^{\theta}} \), which necessitates the application of both the chain rule and the quotient rule to find the derivative properly.

Understanding this concept is central to solving problems that involve logarithmic differentiation in calculus. It's like peeling an onion, where you must carefully unwrap each layer to get to the core solution. The natural logarithm simplifies the process of differentiation when applied correctly, as seen in our example.

Quotient Rule in Calculus

The quotient rule is a fundamental tool in calculus used to differentiate functions that are ratios of two other functions. When you have a function \( y = \frac{v}{w} \), the derivative is not simply the derivative of the top divided by the derivative of the bottom. Instead, calculus provides us with an ingenious formula:

\[ \left( \frac{v}{w} \right)' = \frac{v'w - vw'}{w^2} \]

• \( v \) and \( w \) are the functions on the top (numerator) and bottom (denominator) respectively.
• \( v' \) and \( w' \) are the derivatives of \( v \) and \( w \).

This formula ensures that the derivative accounts for both the change in the numerator and the denominator, which is essential for maintaining the function's behavior across all its points.

In our exercise, \( v = e^\theta \) and \( w = 1 + e^\theta \). Applying the quotient rule gave us \( \frac{e^{\theta}(1 + e^{\theta}) - e^{\theta} e^{\theta}}{(1 + e^{\theta})^2} \), simplifying down to \( \frac{e^{\theta}}{(1 + e^{\theta})^2} \). This step is critical for correctly differentiating composite functions and should be practiced to mastery.

Chain Rule in Calculus

The chain rule is often used in tandem with other differentiation rules like the quotient rule. Its primary use is in computing the derivative of composite functions. When you have a function \( y = f(g(x)) \), the chain rule provides a way to find the derivative \( \frac{dy}{dx} \) by calculating \( f'(g(x)) \cdot g'(x) \). In essence, it "chains" the rate of change from the outer function to the inner function.

In our problem, the argument of the logarithm is a composite function \( u = \frac{e^{\theta}}{1+e^{\theta}} \). To differentiate \( y = \ln(u) \), the chain rule was applied after using the quotient rule to find \( \frac{du}{d\theta} = \frac{e^{\theta}}{(1 + e^{\theta})^2} \).

This step essentially "links" the derivative of the natural logarithm with respect to its immediate argument (\( u \)) to the deeper derivative of \( u \) itself with respect to \( \theta \). The chain rule is a powerful technique, necessary for handling layers of functions dynamically intertwined, and aids in translating complex dependencies into tangible results.