Problem 32 - Thomas Calculus

Question

In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln \left(\frac{\sqrt{\theta}}{1+\sqrt{\theta}}\right) $$

Step-by-Step Solution

Answer

The derivative is \( \frac{1}{2\theta(1+\sqrt{\theta})} \).

Step 1: Apply the Natural Logarithm Rule

To find the derivative of the logarithm function, use the rule: \[ \frac{d}{d\theta} \ln(u) = \frac{1}{u} \cdot \frac{du}{d\theta} \]Here, \( u = \frac{\sqrt{\theta}}{1+\sqrt{\theta}} \).

Step 2: Simplify the Function

Rewrite \( u \) before differentiating: \[ u = \frac{\theta^{1/2}}{1 + \theta^{1/2}} \].

Step 3: Differentiate the Numerator and Denominator

The derivative of the numerator, \( \theta^{1/2} \), is \[ \frac{1}{2}\theta^{-1/2} = \frac{1}{2\sqrt{\theta}} \].The derivative of the denominator, \( 1 + \theta^{1/2} \), is\[ \frac{1}{2\sqrt{\theta}} \].

Step 4: Apply the Quotient Rule

Using the quotient rule: \[ \frac{d}{d\theta} \left( \frac{f}{g} \right) = \frac{f'g - fg'}{g^2} \]where \( f = \theta^{1/2} \), \( g = 1 + \theta^{1/2} \), \( f' = \frac{1}{2\sqrt{\theta}} \), and \( g' = \frac{1}{2\sqrt{\theta}} \).

Step 5: Calculate the Derivative of u

Substitute into the quotient rule:\[\frac{d}{d\theta} \left( \frac{\theta^{1/2}}{1+\theta^{1/2}} \right) = \frac{\left( \frac{1}{2\sqrt{\theta}} \right)(1+\theta^{1/2}) - \left( \frac{1}{2\sqrt{\theta}} \right)\theta^{1/2}}{(1+\theta^{1/2})^2}\]Which simplifies to \[ \frac{1}{2\sqrt{\theta}(1+\theta^{1/2})^2} \].

Step 6: Substitute Back into the Logarithmic Derivative

Substitute the result from Step 5 into the log derivative:\[\frac{d}{d\theta} \ln \left( \frac{\sqrt{\theta}}{1+\sqrt{\theta}} \right) = \frac{1}{\frac{\sqrt{\theta}}{1+\sqrt{\theta}}} \cdot \frac{1}{2\sqrt{\theta}(1+\theta^{1/2})^2}\]Simplify to \[ \frac{1+\sqrt{\theta}}{2\theta(1+\sqrt{\theta})^2} \]Which reduces to \[ \frac{1}{2\theta(1+\sqrt{\theta})} \].

Key Concepts

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a fundamental function in calculus that has many applications in differentiation and integration. When dealing with the derivative of a function that includes a natural logarithm, the chain rule often comes into play. This rule states that if \( y = \ln(u) \), then the derivative \( \frac{dy}{d\theta} \) can be found using the formula:

• \( \frac{d}{d\theta} \ln(u) = \frac{1}{u} \cdot \frac{du}{d\theta} \)

To differentiate \( \ln \left( \frac{\sqrt{\theta}}{1+\sqrt{\theta}} \right) \), we first consider the inner function \( u = \frac{\sqrt{\theta}}{1+\sqrt{\theta}} \). By applying the formula, the derivative of \( y \) in terms of \( \theta \) involves finding the derivative of \( u \) first and then plugging it into the chain rule. This approach helps in breaking down the problem into simpler parts, making the differentiation more manageable.

Quotient Rule

The quotient rule is a handy differentiation technique applied when you are given a function that is the quotient of two other functions. This rule is expressed as:

• \( \frac{d}{d\theta} \left( \frac{f}{g} \right) = \frac{f'g - fg'}{g^2} \)

In the given function \( u = \frac{\sqrt{\theta}}{1+\sqrt{\theta}} \), both the numerator and denominator involve differentiable expressions. Here, the numerator \( f = \sqrt{\theta} \) and denominator \( g = 1 + \sqrt{\theta} \) both need to be differentiated independently:

• \( f' = \frac{1}{2\sqrt{\theta}} \)
• \( g' = \frac{1}{2\sqrt{\theta}} \)

Then, applying the quotient rule, we substitute \( f \), \( g \), \( f' \), and \( g' \) into the formula, to find the derivative of \( u \). This step is crucial as it prepares us to plug back this result into the natural logarithm's chain rule.

Simplification

Simplifying complex expressions is not just crucial for clarity but also for obtaining the final derivative in the simplest form possible. Once the derivative of \( u \) has been calculated using the quotient rule, the next task involves simplifying the resulting expression. Initially, we find:

• \( \frac{d}{d\theta} \left( \frac{\theta^{1/2}}{1+\theta^{1/2}} \right) = \frac{1}{2\sqrt{\theta}(1+\theta^{1/2})^2} \)

Upon substitution into the logarithmic differentiation formula, pay attention to eliminating complex fractions, common factors, or any cancelable terms. Simplifying to \( \frac{1}{2\theta(1+\sqrt{\theta})} \) showcases the power of simplification, transforming complex steps into an easy-to-read result. This step demonstrates the elegance of mathematical expression and the critical role of simplification in calculus.

Differentiation with Respect to Theta

Differentiation generally means finding the rate at which a function changes as its input changes. In this exercise, differentiation is performed with respect to \( \theta \), a variable often used in trigonometry and polar coordinates. The process involves applying known rules to simplify functions of \( \theta \). When differentiating a composite function like \( y = \ln \left( \frac{\sqrt{\theta}}{1+\sqrt{\theta}} \right) \), it requires applying the chain rule first, followed by particular rules like the quotient rule for different parts of \( u \).Through differentiation, we aim to express how \( y \) changes with slight changes in \( \theta \). This requires understanding how each component of the function behaves. Hence, each minor derivative concerning \( \theta \) contributes cumulatively to determine how the overall function reacts when \( \theta \) is varied, reflecting the fundamental concept of rates of change in calculus.