Problem 67 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the given independent variable. $$y=\log _{2} 5 \theta$$

Step-by-Step Solution

Answer

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

Step 1: Review the Logarithm Property

The function given is a logarithm with base 2. We have the function \( y = \log_{2}(5\theta) \). Using the property of logarithms, we can rewrite this as \( y = \frac{\ln (5\theta)}{\ln 2} \). This step leverages the conversion from any logarithm base to the natural logarithm \( \ln \), which can simplify differentiation.

Step 2: Apply the Chain Rule

To find the derivative of \( y \) with respect to \( \theta \), we'll apply the chain rule to \( y = \frac{\ln (5\theta)}{\ln 2} \). The chain rule states that \( \frac{d}{d\theta}[u(v)] = u'(v) \cdot v'(\theta) \). Here, \( u(v) = \ln(v) \) and \( v = 5\theta \).

Step 3: Differentiate with Respect to \( \theta \)

We take the derivative of \( \ln (5\theta) \) using the chain rule. The derivative of \( \ln(5\theta) \) with respect to \( \theta \) is \( \frac{1}{5\theta} \times 5 = \frac{5}{5\theta} = \frac{1}{\theta} \).

Step 4: Simplify the Derivative Expression

Since \( y' = \frac{1}{\theta} \times \frac{1}{\ln 2} \), the derivative \( \frac{dy}{d\theta} = \frac{1}{\theta \ln 2} \). This gives us the final simplified form for the derivative.

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus used for finding the derivative of a composite function. It allows us to differentiate complex functions by breaking them down into their simpler parts. In essence, if you have a function nested inside another, the chain rule helps us differentiate it easily.
Here's how it works:

• Suppose you have a composite function defined as \( y = u(v(x)) \), where \( u \) and \( v \) are differentiable functions.
• The chain rule states that the derivative of \( y \) with respect to \( x \) is \( u'(v(x)) \cdot v'(x) \).

In applying the chain rule, you first differentiate the outer function and then multiply it by the derivative of the inner function. This results in an expression that accurately describes how the function's output changes with respect to the input. In our example, we applied the chain rule to differentiate the function \( \ln(5\theta) \).

Logarithm Properties

Logarithms have several properties that are extremely useful in simplifying and transforming mathematical expressions. One major property that comes handy is how logarithms of different bases can be converted using natural logarithms.
This particular property is expressed as:

• \( \log_b (x) = \frac{\ln(x)}{\ln(b)} \)

This means that you can transform any logarithm into a form that uses the natural logarithm \( \ln \).
For example, to differentiate \( y = \log_2(5\theta) \), we first convert it using the property into: \( y = \frac{\ln(5\theta)}{\ln(2)} \). This transformation eases the differentiation process as it allows us to leverage the differentiation rules of natural logs. Remember, using logarithm properties helps in breaking down complex logs into manageable parts, making calculations simpler.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is a specific logarithm with the base \( e \), where \( e \approx 2.71828 \). It has unique properties that make it very useful in calculus, especially when finding derivatives.
Key characteristics of natural logarithms include:

• \( \ln(1) = 0 \)
• \( \ln(e) = 1 \)
• The derivative: \( \frac{d}{dx}[\ln(x)] = \frac{1}{x} \)

In our exercise, we utilized the natural logarithm to convert \( \log_2(5\theta) \) to \( \frac{\ln(5\theta)}{\ln(2)} \), a form easily differentiable with known derivative formulas. The formula \( \frac{1}{x} \) when differentiating \( \ln(x) \) means that for our function \( \ln(5\theta) \), the derivative become \( \frac{1}{5\theta} \times 5 \), simplifying to \( \frac{1}{\theta} \).

Derivative Calculation

Calculating derivatives is a central concept in calculus that involves finding the rate at which something changes. It's why functions are powerful, as derivatives provide insight into behaviors such as speed, growth, and trends. The basic goal is to find the slope of the tangent line to the curve of a function at any point.
For differentiating functions that involve products or chains, such as in our exercise, applying rules like the chain rule or product rule becomes vital. In our case:

• We first rewrote \( y = \log_2(5\theta) \) as \( \frac{\ln(5\theta)}{\ln(2)} \).
• Using the chain rule, we differentiated \( \ln(5\theta) \) as explained.
• Finally, we simplified the result to get the derivative \( \frac{1}{\theta \ln(2)} \).

Understanding these rules and how to apply them makes tackling more complex functions manageable. Differentiation helps in identifying how small changes in \( \theta \) affect \( y \), giving precise control and understanding over the function's behavior.