Problem 10 - Thomas Calculus

Question

In Exercises \(5-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln \frac{10}{x} $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{1}{x}.\)

Step 1: Express the Function Differently

First, rewrite the function \( y = \ln \frac{10}{x} \) using the properties of logarithms. We can express it as \( y = \ln 10 - \ln x \). This simplification will make differentiating easier.

Step 2: Differentiate the Function

Differentiate each term with respect to \( x \). The derivative of a constant, \( \ln 10 \), is 0. For \( \ln x \), the derivative is \( \frac{1}{x} \). Thus, \( \frac{dy}{dx} = 0 - \frac{1}{x} \).

Step 3: Write the Final Result

Combine the differentiated terms to find the derivative of the entire function. The resulting derivative is \( \frac{dy}{dx} = -\frac{1}{x} \).

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a useful technique, especially when dealing with complicated expressions or products and quotients within a function. Here, we have to differentiate functions that involve logarithms, which can sometimes seem intimidating. In this scenario, you can leverage the properties of logarithms to simplify the function before differentiation. By converting products, quotients, or powers into simpler additive or subtractive terms using logarithms, we can make the differentiation process easier. Basically, what we do is take the natural logarithm of both sides of an equation if needed, which often breaks down complex expressions into more manageable parts. Then, after simplifying, we can differentiate as usual. For the given function, logarithmic differentiation helps to break down the quotient inside the logarithm into two separate terms, thanks to the properties of logs. Always remember: logarithmic differentiation can be a go-to technique when dealing with intricate functions that involve multiplication or division.

Properties of Logarithms

The properties of logarithms play a pivotal role in transforming functions into forms that are easier to differentiate. These properties simplify expressions by decomposing complex operations into simpler ones. Here are the main properties that are frequently used in calculus:

• Product Rule: \ \ \( \log_b(MN) = \log_b(M) + \log_b(N) \)
• Quotient Rule: \ \ \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \)
• Power Rule: \ \ \( \log_b(M^p) = p\log_b(M) \)

In our example, we utilize the quotient rule of logarithms to rewrite the initial expression from \( y = \ln \frac{10}{x} \) to \( y = \ln 10 - \ln x \). This transformation is crucial because it allows one to handle the differentiation in a piece-by-piece manner rather than dealing with the original compound form.Remembering these properties and being comfortable applying them will greatly enhance your ability to tackle logarithmic differentiation problems effectively and efficiently.

Differentiation Techniques

Differentiation techniques are the backbone of calculus and allow us to find the rates at which things change. With the simplification from logarithm properties, we use basic differentiation rules to find derivatives effortlessly. The most common rules used in differentiation include:

• Constant Rule: The derivative of a constant is zero.
• The Power Rule: The derivative of \( x^n \) is \( nx^{n-1} \).
• The Chain Rule: Used when differentiating the composition of functions.
• The Product and Quotient Rules: Used for functions that are products or quotients of other functions.

In our problem, after employing the properties of logarithms, we utilize the constant rule on \( \ln 10 \), which is why its derivative is 0. For the term \( \ln x \), we apply the standard logarithmic differentiation rule, which states that the derivative of \( \ln x \) is \( \frac{1}{x} \).In the end, combining these simple applications of differentiation rules gives us the final derivative \( \frac{dy}{dx} = -\frac{1}{x} \). Understanding and applying these strategies efficiently will strengthen your grasp of calculus and ultimately simplify complex differentiations.