Problem 60 - University Calculus: Early Transcendentals

Question

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

\(\frac{dy}{d\theta} = \frac{1}{2\theta} - \frac{1}{2\sqrt{\theta}(1 + \sqrt{\theta})}\).\)

Step 1: Recognize the Form

The function is given as a logarithm. To find the derivative, we'll use the chain rule combined with properties of logarithms. We'll start by simplifying the expression inside the logarithm.

Step 2: Use Logarithm Properties

We know that \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \). Applying this to our function, we get:\[ y = \ln(\sqrt{\theta}) - \ln(1 + \sqrt{\theta}) \]

Step 3: Simplify using Logarithm Rules

We apply the logarithm rule \( \ln(a^b) = b \ln(a) \) to the terms:\[ y = \frac{1}{2} \ln(\theta) - \ln(1 + \sqrt{\theta}) \]

Step 4: Differentiate Each Part

We differentiate each part separately. The derivative of \( \frac{1}{2} \ln(\theta) \) with respect to \( \theta \) is \( \frac{1}{2\theta} \), and the derivative of \( \ln(1 + \sqrt{\theta}) \) involves a chain rule: \( \frac{d}{d\theta} \ln(1 + \sqrt{\theta}) = \frac{1}{1 + \sqrt{\theta}} \times \frac{1}{2\sqrt{\theta}} \).

Step 5: Combine the Derivatives

Combine the derivatives to find the overall derivative:\[ \frac{dy}{d\theta} = \frac{1}{2\theta} - \frac{1}{2\sqrt{\theta}(1 + \sqrt{\theta})} \].

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a useful technique when working with complex exponentials or products in a function. It involves taking the natural logarithm of both sides of an equation to simplify the differentiation process. By converting multiplication into addition and division into subtraction, it makes differentiation more manageable. In this exercise, we take the logarithm of the expression given by \( y = \ln \left( \frac{\sqrt{\theta}}{1 + \sqrt{\theta}} \right) \).
After applying the properties of logarithms, this becomes the difference between two simpler logarithms: \( \ln(\sqrt{\theta}) - \ln(1 + \sqrt{\theta}) \).
This step prepares us for straightforward differentiation by transforming the division into subtraction.

Chain Rule

The chain rule is essential when dealing with composite functions, which are functions nested within other functions. It states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).
In our exercise, the expression \( \ln(1 + \sqrt{\theta}) \) requires the use of the chain rule because it contains a function inside of another function.
First, differentiate the outer function: the natural log function \( \ln(x) \), which gives us \( \frac{1}{1 + \sqrt{\theta}} \).
Next, differentiate the inner function \( 1 + \sqrt{\theta} \), giving \( \frac{1}{2\sqrt{\theta}} \).
By multiplying these results, we achieve the derivative of the original composite function.

Properties of Logarithms

Logarithmic properties are integral in simplifying expressions and functions, especially prior to differentiation. They transform multiplication into addition and division into subtraction, making it easier to perform calculus operations.
Key properties used in this exercise include:

• The quotient rule for logarithms: \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)
• The power rule for logarithms: \( \ln(a^b) = b \cdot \ln(a) \)

Applied to our original equation, these rules allow us to break the expression into manageable parts. We simplify \( \ln(\sqrt{\theta}) \) to \( \frac{1}{2} \ln(\theta) \) using the power rule.
This prepares the equation for differentiation, turning it into a more digestible format.

Differentiation Techniques

Differentiation is a crucial process in calculus, used for finding how a function changes at any point. This exercise involves advanced differentiation techniques.
For example, differentiating \( \frac{1}{2} \ln(\theta) \) with respect to \( \theta \) results in \( \frac{1}{2\theta} \), derived from the derivative of \( \ln(\theta) \) being \( \frac{1}{\theta} \).
Similarly, the application of the chain rule allows us to find the derivative of \( \ln(1 + \sqrt{\theta}) \). These techniques map out the path to conveniently isolate and differentiate each part of the function.
Finally, combining these derivative results gives the full derivative of the original function: \( \frac{dy}{d\theta} = \frac{1}{2\theta} - \frac{1}{2\sqrt{\theta}(1 + \sqrt{\theta})} \).
Mastery of these techniques ensures the ability to tackle complex differentiation problems efficiently.