Problem 45 - University Calculus: Early Transcendentals

Question

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=(\sin \theta) \sqrt{\theta+3}$$

Step-by-Step Solution

Answer

The derivative is \(\frac{dy}{d\theta} = \cos \theta \sqrt{\theta+3} + \frac{(\sin \theta)}{2 \sqrt{\theta+3}}\).

Step 1: Take the Logarithm of Both Sides

To use logarithmic differentiation, start by taking the natural logarithm of both sides of the equation. This is: \[ \ln(y) = \ln \left( (\sin \theta) \sqrt{\theta + 3} \right) \] Use the property of logarithms to separate the product: \[ \ln(y) = \ln(\sin \theta) + \ln(\sqrt{\theta + 3}) \] Since \(\ln(\sqrt{\theta + 3}) = \frac{1}{2} \ln(\theta + 3)\), we have: \[ \ln(y) = \ln(\sin \theta) + \frac{1}{2} \ln(\theta + 3) \]

Step 2: Differentiate Both Sides

Differentiate both sides of the equation with respect to \(\theta\). For the left side, using the chain rule: \[ \frac{1}{y} \frac{dy}{d\theta} \] For the right side, differentiate term by term:1. \( \frac{d}{d\theta} \ln(\sin \theta) = \cot \theta \)2. \( \frac{1}{2} \cdot \frac{d}{d\theta} \ln(\theta + 3) = \frac{1}{2} \cdot \frac{1}{\theta + 3} \)Thus, \[ \frac{1}{y} \frac{dy}{d\theta} = \cot \theta + \frac{1}{2(\theta + 3)} \]

Step 3: Solve for the Derivative

To solve for the derivative \( \frac{dy}{d\theta} \), multiply both sides by \(y\): \[ \frac{dy}{d\theta} = y \left( \cot \theta + \frac{1}{2(\theta + 3)} \right) \]Substitute back \( y = (\sin \theta) \sqrt{\theta + 3} \):\[ \frac{dy}{d\theta} = (\sin \theta) \sqrt{\theta + 3} \left( \cot \theta + \frac{1}{2(\theta + 3)} \right) \]

Step 4: Simplify the Expression

Simplify the expression, using \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), then: \[ \cot \theta \cdot \sin \theta = \cos \theta \]Thus, \[ \frac{dy}{d\theta} = (\sin \theta) \sqrt{\theta + 3} \left( \frac{\cos \theta}{\sin \theta} + \frac{1}{2(\theta + 3)} \right) \]Simplify further to get: \[ \frac{dy}{d\theta} = \cos \theta \sqrt{\theta + 3} + \frac{(\sin \theta)}{2 \sqrt{\theta + 3}} \]

Key Concepts

Chain Rule

The Chain Rule is a very useful tool when it comes to taking derivatives, especially in situations where you have a function inside a function. It's like peeling away layers of an onion! If you have a composition of functions, say \( h(x) = f(g(x)) \), then to find \( h'(x) \), you apply the chain rule:
\[ h'(x) = f'(g(x)) imes g'(x). \]
In our logarithmic differentiation case, when we take the derivative of both sides of \( \ln(y) \), we find:

• The left side becomes \( \frac{1}{y} \frac{dy}{d\theta} \), due to the chain rule.
• This only works because \( y \) is a function of another variable, \( \theta \).

You differentiate the outer function (the logarithm) and then multiply by the derivative of the inner function (\( y \)), as per the chain rule's guidelines.
This step is crucial as it allows us to eventually solve for \( \frac{dy}{d\theta} \), by isolating it in the equation.

Logarithm Properties

Logarithm properties are your best friends during logarithmic differentiation! They help simplify complex expressions into manageable pieces. In the problem, using the property:

• \( \ln(ab) = \ln a + \ln b \)

allows us to break down the expression \( \ln \left( (\sin \theta) \sqrt{\theta + 3} \right) \) into smaller parts:
\[ \ln(y) = \ln(\sin \theta) + \ln(\sqrt{\theta + 3}) \]Moreover, knowing that \( \ln(a^b) = b \ln a \) enabled us to write:

• \( \ln(\sqrt{\theta + 3}) = \frac{1}{2} \ln(\theta + 3). \)

By applying these properties, we simplify derivatives calculation significantly.
This transformation makes applying differentiation rules much easier.

Product Rule

The Product Rule is essential when dealing with derivatives of products. It states that if you have two functions, \( u(x) \) and \( v(x) \), then the derivative of their product \( u(x)v(x) \) is
\[ \frac{d}{dx}(uv) = u'v + uv'. \]
However, in the context of logarithmic differentiation, we cleverly avoid having to directly use this rule by transforming products into sums with logarithms.
Through the properties of logarithms, the difficult task of differentiating a product directly is circumvented, making our job simpler.
Instead of applying the product rule in the traditional sense, we consolidate these expressions using logarithm properties, saving time and avoiding potential errors.
In our problem, this method enables us to manage the differentiation efficiently by breaking the initial product under the log into separate terms which we can differentiate individually.