Problem 78 - Thomas Calculus in SI Units

Question

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

Step-by-Step Solution

Answer

The derivative of the given function using the quotient rule is computed and simplified for the final expression.

Step 1: Identify the Derivative Rule

The function is a quotient, so we'll use the quotient rule for derivatives. The quotient rule states that if you have a function in the form \( y = \frac{u}{v} \), then the derivative \( y' \) is \( \frac{u'v - uv'}{v^2} \).

Step 2: Define the Functions

Identify \( u \) and \( v \) in the function \( y = \frac{u}{v} \). Here, \( u = \theta 5^{\theta} \) and \( v = 2 - \log_5 \theta \).

Step 3: Differentiate the Numerator

Differentiate \( u = \theta 5^{\theta} \) with respect to \( \theta \). Use the product rule \( (uv)' = u'v + uv' \), where \( u = \theta \) and \( v = 5^{\theta} \). So, \( u' = 1 \) and \( v' = 5^{\theta} \ln(5) \). Thus, \( u' = 5^{\theta} + \theta \cdot 5^{\theta} \ln(5) \).

Step 4: Differentiate the Denominator

Differentiate \( v = 2 - \log_5 \theta \) using the formula for the derivative of a logarithm. Recall that \( \log_b x' = \frac{1}{x \ln(b)} \), so \( v' = -\frac{1}{\theta \ln(5)} \).

Step 5: Apply the Quotient Rule

Plug \( u, u', v, \) and \( v' \) into the quotient rule formula: \( y' = \frac{(5^{\theta} + \theta \cdot 5^{\theta} \ln(5))(2 - \log_5 \theta) - (\theta 5^{\theta})(-\frac{1}{\theta \ln(5)})}{(2 - \log_5 \theta)^2} \).

Step 6: Simplify the Expression

Complete the computation of the derivative by simplifying the terms from Step 5. Carefully distribute and combine like terms where possible to find the most concise form of \( y' \).

Key Concepts

Quotient Rule

The Quotient Rule is a vital concept used in calculus, especially when differentiating a function that is a division of two other functions. It allows you to find the derivative of a quotient. If a function is of the form \( y = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable, the derivative \( y' \) is given by:
\[ y' = \frac{u'v - uv'}{v^2}. \]
This formula might look complex at first, but it hinges on relatively simple arithmetic and differentiation rules. The core idea is to take the derivative of the numerator and the denominator separately. You must remember to subtract the product \( u'v \) from \( uv' \), making sure to divide everything by \( v^2 \). This rule is imperative because it accounts for changes in both parts of the quotient, making your derivative accurate.

Product Rule

The Product Rule is another key differentiation rule. It helps when you have two functions that are multiplied together, like \( u = \theta \cdot 5^{\theta} \) in our example. The rule states that the derivative of a product \( uv \) is:
\[ (uv)' = u'v + uv'. \]
This is simple yet powerful because it tells you to differentiate each function one at a time, treating the other as a constant, and then combine the results. You add the two resulting expressions together. It's essential to apply this rule correctly because overlooking one part may lead to an incomplete derivative. This rule is useful whenever differentiating compound products that naturally occur in many functions.

Logarithmic Derivatives

When dealing with logarithmic functions and their derivatives, you use logarithmic differentiation. This allows you to simplify and find the derivative of functions involving logarithms. The fundamental rule here is:
\[ \left(\log_b x\right)' = \frac{1}{x \ln(b)}. \]
Understanding this helps to tackle expressions that involve logarithms, like \( v = 2 - \log_5 \theta \) in our exercise. When differentiating, make sure you understand the base's implication, as it affects the result. This rule simplifies into general formulas that involve logs, greatly assisting in finding derivatives easily and effectively.

Simplification of Expressions

Simplifying expressions is crucial when you have derived a complex form of an expression. After applying differentiation rules and initially computing the derivative, it might look bulky or convoluted. Simplification includes distributing terms where necessary, combining like terms, and using arithmetic rules to condense expressions down to their most understandable form. This final step helps in making the derived formula simpler to work with or interpret. Always aim for expressions that are not only correct but also clear and elegant. This helps in subsequent steps of any problem-solving process, ensuring you and others can follow the thought process without unnecessary complexity.