Problem 21 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{d\theta} = \frac{1}{1+e^{\theta}} \).

Step 1: Identify the Derivative Rule to Use

The function given is a natural logarithm, so we'll use the chain rule in combination with the derivative of the natural logarithm. If \( y = \ln(u) \), then \( \frac{dy}{d\theta} = \frac{1}{u} \cdot \frac{du}{d\theta} \).

Step 2: Define the Inside Function u

Identify the inside function \( u = \frac{e^{\theta}}{1+e^{\theta}} \). We will need to find \( \frac{du}{d\theta} \).

Step 3: Apply the Quotient Rule to Find du/dθ

Use the quotient rule, \( \frac{d}{d\theta}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2} \), to differentiate \( u \).Let \( f = e^{\theta} \) and \( g = 1+e^{\theta} \).Then \( f' = e^{\theta} \) and \( g' = e^{\theta} \).So, \( \frac{du}{d\theta} = \frac{e^{\theta}(1+e^{\theta}) - e^{\theta}e^{\theta}}{(1+e^{\theta})^2} \).

Step 4: Simplify the Derivative of u

Calculate \( \frac{du}{d\theta} = \frac{e^{\theta} + e^{2\theta} - e^{2\theta}}{(1+e^{\theta})^2} = \frac{e^{\theta}}{(1+e^{\theta})^2} \).

Step 5: Use the Chain Rule to Find dy/dθ

Substitute \( u \) and \( \frac{du}{d\theta} \) into the chain rule: \( \frac{dy}{d\theta} = \frac{1}{\frac{e^{\theta}}{1+e^{\theta}}} \cdot \frac{e^{\theta}}{(1+e^{\theta})^2} \).

Step 6: Simplify dy/dθ

Simplifying gives \( \frac{dy}{d\theta} = \frac{1+e^{\theta}}{e^{\theta}} \cdot \frac{e^{\theta}}{(1+e^{\theta})^2} = \frac{1}{1+e^{\theta}} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, designed to handle derivatives of composite functions. Whenever you have a function inside another function, the chain rule allows you to find the derivative of the entire composition without tearing your hair out. Essentially, if you have a composite function, such as when you express it as \( y = f(g(x)) \), the chain rule states: \[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\] Think of it as peeling an onion; you differentiate the outer layer, then proceed to the inner layer. In our exercise, we dealt with the natural logarithm function, \( \ln(u) \), wherein the inside function \( u = \frac{e^{\theta}}{1+e^{\theta}} \). We used the chain rule to differentiate \( \ln(u) \), following it by the derivative of \( u \). Always remember the chain rule's secret weapon is its ability to systematically tackle functions with layers, just like an onion. 😊 Not only is it efficient, but it's also quite elegant!

Quotient Rule

The quotient rule is another tool in the calculus toolkit for finding the derivative of a quotient of two functions. When you're navigating a function like \( \frac{f(x)}{g(x)} \), the quotient rule helps you calculate its derivative with ease. The formula for the quotient rule is: \[\frac{d}{dx} \left( \frac{f}{g} \right) = \frac{f'g - fg'}{g^2}\] Here, \( f \) is the numerator, and \( g \) is the denominator. This rule comes in handy in our exercise when we differentiate \( u = \frac{e^{\theta}}{1+e^{\theta}} \). To apply the quotient rule:

• Differentiate the top function (\( f \)) and bottom function (\( g \)). In our example, both the numerator and denominator differentiate to \( e^{\theta}\).
• Use the quotient rule formula to get the derivative of the fraction.

The beauty of the quotient rule lies in its ability to quickly compute the derivative of a fraction, safeguarding against potential rookie errors.

Natural Logarithm Differentiation

Differentiating natural logarithms can seem daunting, but it's straightforward once you grasp the rule. The derivative of a natural logarithm \( \ln(x) \) is simply \( \frac{1}{x} \). When dealing with more complex arguments, like a function \( u(x) \) inside the logarithm, it combines neatly with the chain rule. If \( y = \ln(u) \), the derivative \( \frac{dy}{dx} \) becomes: \[\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}\] In our example from the problem, the natural logarithm \( \ln \left( \frac{e^{\theta}}{1+e^{\theta}} \right) \) required us to derive the inside function \( u \) first. We then combined that result with the rule for differentiating natural logs to find \( \frac{dy}{d\theta} \). This process highlights the elegance and simplicity of using the natural logarithm differentiation rule with the chain rule. Breaking down the steps is essential, making sure to understand each layer before moving on. With practice, differentiating logarithmic functions becomes a naturally enjoyable process!