Problem 56 - Calculus Single Variable

Question

In Exercises \(56-59\), evaluate the derivative \(f^{\prime}\) of the given function \(f\) in two ways. First, apply the Chain Rule to \(f(x)\) without simplifying \(f(x)\) in advance. Second, simplify \(f(x)\), and then differentiate the simplified expression. Verify that the two expressions are equal. $$ f(x)=\ln (3 x) $$

Step-by-Step Solution

Answer

The derivative of \(f(x) = \ln (3x)\) is \(f'(x) = \frac{1}{x}\), verified by both methods.

Step 1: Differentiate without Simplifying

To differentiate \(f(x) = \ln (3x)\) using the Chain Rule, we let \(u = 3x\), so \(f(x) = \ln u\). The derivative of \(\ln u\) with respect to \(u\) is \(\frac{1}{u}\). Additionally, the derivative of \(u = 3x\) with respect to \(x\) is \(3\). Apply the Chain Rule by multiplying these derivatives: \[ f'(x) = \frac{1}{3x} \cdot 3 = \frac{3}{3x} = \frac{1}{x}.\]

Step 2: Simplify the Function

Simplify \(f(x) = \ln (3x)\) using the properties of logarithms: \[f(x) = \ln 3 + \ln x.\]

Step 3: Differentiate the Simplified Form

Now, differentiate \(f(x) = \ln 3 + \ln x\). The derivative of a constant \(\ln 3\) is zero, and the derivative of \(\ln x\) is \(\frac{1}{x}\). Therefore, the derivative is: \[f'(x) = 0 + \frac{1}{x} = \frac{1}{x}.\]

Step 4: Verify Equality of Results

Compare the derivatives obtained from Step 1 and Step 3. Both methods yield the same derivative, \(f'(x) = \frac{1}{x}\), verifying the correctness of the calculations.

Key Concepts

Chain Rule

Understanding the Chain Rule is essential when handling more complex functions in calculus, especially those involving compositions of functions. The Chain Rule enables us to find the derivative of a composite function.

Here’s a step-by-step breakdown to help grasp this concept:

• A composite function is essentially a function within a function. In our example, the function \(f(x) = \ln(3x)\) is composed of \(u = 3x\) inside the natural logarithm function.
• The Chain Rule states that to find \(f'(x)\), the derivative of a composite function, we take the derivative of the outer function with respect to the inner function, then multiply it by the derivative of the inner function with respect to x.
• In notation form, if \(y = f(g(x))\), then \(y' = f'(g(x)) \cdot g'(x)\).
• By applying the Chain Rule, we worked out \(f'(x) = \frac{1}{u} \cdot 3 = \frac{3}{3x} = \frac{1}{x}\).

This rule is powerful as it can be applied to a variety of functions, making it a crucial tool in your calculus toolkit.

Derivatives

Derivatives are a fundamental aspect of calculus that describe how a function changes at any point.

They tell us about the rate of change or the slope of the function at a given point. In our example, the derivative \(f '(x) = \frac{1}{x}\) describes how the function \(f(x) = \ln(3x)\) changes as \(x\) changes.

• Essentially, the derivative provides the velocity of the function’s output as the input grows or shrinks.
• It lets us know whether the function is increasing or decreasing and how steep the curve is.
• For \(f(x) = \ln(3x)\), the derivative indicates that as \(x\) increases, \(f(x)\) rates of change are inversely proportional.
• Remember, the derivative of \(\ln x\) is \(\frac{1}{x}\), which comes handy when dealing with logarithmic functions.

Understanding derivatives helps in understanding real-world problems requiring measurements of change, such as physics or economics.

Logarithmic Differentiation

Logarithmic Differentiation is a special technique in calculus that simplifies the process of finding the derivative of more complex functions, especially those that involve products or quotients.

This method uses the properties of logarithms to transform difficult multiplication or division in functions into easier addition or subtraction operations.

• In our problem, simplifying \(f(x) = \ln(3x)\) to \(\ln 3 + \ln x\) allowed us to differentiate each term separately, reducing the complexity.
• The key property used is \(\ln(a \cdot b) = \ln a + \ln b\), helping to break down products into sums.
• After simplification, other derivative rules can be easily applied, as seen with \(\frac{1}{x}\) being the derivative of \(\ln x\).
• Logarithmic Differentiation proves particularly useful when working with products of functions and when exponentiation is involved, owing to its simplicity.

By leveraging logarithmic differentiation, complex problems become manageable, making it a valuable technique in advanced calculus.