Problem 74 - Calculus Early Transcendentals

Question

Alternative proof of product property Assume that \(y>0\) is fixed and that \(x>0 .\) Show that \(\frac{d}{d x}(\ln x y)=\frac{d}{d x}(\ln x) .\) Recall that if two functions have the same derivative, they differ by an additive constant. Set \(x=1\) to evaluate the constant and prove that \(\ln x y=\ln x+\ln y.\)

Step-by-Step Solution

Answer

Question: Prove the product property of logarithms \(\ln(xy) = \ln(x) + \ln(y)\) using an alternative method, where \(x>0\) and \(y>0\) is fixed.

Step 1: Find the derivative of \(\ln(xy)\) with respect to \(x\)

Since \(y\) is a constant, use the chain rule to find the derivative of \(\ln(xy)\) with respect to \(x\). The chain rule in this context states that the derivative of a function composed of two functions is the product of the derivative of the outer function and the derivative of the inner function. In this case, the outer function is \(\ln(u)\) and the inner function is \(u = xy\). First, find the derivative of the outer function with respect to \(u\) (denote this as \(\frac{d\ln(u)}{du}\)): \(\frac{d\ln(u)}{du} = \frac{1}{u}\) Next, find the derivative of the inner function with respect to \(x\) (denote this as \(\frac{d(xy)}{dx}\)): \(\frac{d(xy)}{dx} = y\) Now, apply the chain rule: \(\frac{d(\ln(xy))}{dx} = \frac{d\ln(u)}{du} \cdot \frac{d(xy)}{dx} = \frac{1}{u} \cdot y = \frac{y}{xy} = \frac{1}{x}\)

Step 2: Calculate the derivative of \(\ln(x)\) with respect to \(x\)

To find the derivative of \(\ln(x)\) with respect to \(x\), simply differentiate the natural logarithm function: \(\frac{d(\ln(x))}{dx} = \frac{1}{x}\)

Step 3: Compare the derivatives

Since both \(\frac{d(\ln(xy))}{dx}\) and \(\frac{d(\ln(x))}{dx}\) are equal to \(\frac{1}{x}\), these functions differ by an additive constant, say \(c\).

Step 4: Determine the constant value, \(c\)

To determine the constant, we can set \(x=1\). \(\ln(1y) = \ln(1) + \ln(y) + c\) Since \(\ln(1) = 0\): \(\ln(y) = \ln(y) + c\) Therefore, the constant \(c = 0\).

Step 5: Prove the product property

Since the constant value \(c=0\) and the derivative of \(\ln(xy)\) is equal to the derivative of \(\ln(x)\), the product property holds true: \(\ln(xy) = \ln(x) + \ln(y)\)

Key Concepts

Chain Rule in Calculus

Calculus involves many rules that help us differentiate functions, and one essential rule is the Chain Rule. The Chain Rule is vital when dealing with composite functions, where one function is nested inside another. It allows us to find the derivative of such functions effectively.
For the function \(ln(xy)\), we treat it as a composition of the outer function \(ln(u)\) and the inner function \(u = xy\). According to the Chain Rule, the derivative of this composite function is found by multiplying the derivative of the outer function by the derivative of the inner function.

• The derivative of the outer function, \(ln(u)\), with respect to \(u\) is \(\frac{1}{u}\). This means as \(u\) changes, the rate at which \(ln(u)\) changes is given by this derivative.
• The derivative of the inner function, \(xy\), with respect to \(x\) is \(y\), because \(y\) is treated as a constant during differentiation.

Putting it all together, the Chain Rule tells us that the derivative of \(ln(xy)\) is \(\frac{1}{u} \cdot y = \frac{y}{xy} = \frac{1}{x}\). This shows how powerful the Chain Rule can be when tackling problems involving composite functions.

Natural Logarithm Derivative

The natural logarithm, denoted as \(ln(x)\), is a fundamental function in calculus. It has a simple but powerful derivative that shows how the logarithm function changes with respect to \(x\).
The derivative of \(ln(x)\) is \(\frac{1}{x}\), illustrating that as \(x\) increases, the rate of change of \(ln(x)\) decreases, but never becomes negative.
This property makes the natural logarithm particularly useful in many areas of mathematics and science.

• When differentiating \(ln(x)\), we find that the result is solely dependent on \(x\), allowing us to apply this knowledge directly to various problems involving logarithmic functions.
• Due to its straightforward nature, the derivative helps simplify the computations involved when working with more complex functions that include logarithms.

The constant derivative value \(\frac{1}{x}\) plays an essential role in confirming mathematical properties like the product property, as seen in the exercise to prove \(ln(xy) = \ln(x) + \ln(y)\).

Mathematical Proof

Mathematical proofs are the cornerstone of establishing truth and certainty in mathematics. They help demonstrate why a mathematical statement is correct and provide the logical steps that validate a concept or rule.
In the exercise, the task was to prove the product property of logarithms, \(ln(xy) = \ln(x) + \ln(y)\).
The proof involves:

• First, finding the derivatives of relevant functions using calculus (in this case for \(ln(xy)\) and \(ln(x)\)).
• Second, showing that these derivatives are equal, indicating that their difference is a constant, \(c\).
• Third, evaluating the functions at a specific point (using \(x = 1\)) to find the value of \(c\).
• Finally, concluding that since the constant is zero, \(ln(xy)\) and \(ln(x) + \ln(y)\) are equal.

This structured approach not only proves the property but also deepens understanding of why it holds true. Developing the ability to construct such proofs is crucial for any student of mathematics, as it enhances logical reasoning and problem-solving skills.