Problem 55
Question
Let \(f\) be a differentiable function that is defined on \(\mathbb{R}^{+}\) and that satisfies identity \(f(x y)=f(x)+f(y) .\) This exercise outlines a proof that $$ f(x)=f^{\prime}(1) \cdot \ln (x) $$ a. By differentiating both sides of the given equation with respect to \(x,\) treating \(y\) as constant, show that \(f^{\prime}(x y) \cdot y=f^{\prime}(x)\) b. By setting \(x=t\) and \(y=1 / t\) in the equation obtained in part a, show that \(f\) is an antiderivative of the function \(t \mapsto f^{\prime}(1) / t\) c. Use the result of part \(\mathrm{c}\) to deduce that \(f(t)=f^{\prime}(1)\). \(\ln (t)+C\) where \(C\) is a constant. By evalutating \(f(t)\) at \(t=1,\) show that \(C=f(1)\) d. By considering the given property of \(f\) with \(x=y=1\), deduce that \(f(1)=0\)
Step-by-Step Solution
Verified Answer
The solution is \( f(x) = f'(1) \ln(x) \), with \( f(1) = 0 \).
1Step 1: Differentiate f(xy) with respect to x
Given the equation \( f(xy) = f(x) + f(y) \), we differentiate both sides with respect to \( x \), treating \( y \) as a constant. By applying the chain rule on the left side, we obtain:\[ \frac{d}{dx}[f(xy)] = f'(xy) \cdot y \]On the right side, we differentiate \( f(x) + f(y) \) to get:\[ f'(x) \]Setting these two expressions equal gives us:\[ f'(xy) \cdot y = f'(x) \]
2Step 2: Set x = t and y = 1/t
For the expression \( f'(xy) \cdot y = f'(x) \) obtained in Step 1, substitute \( x = t \) and \( y = 1/t \):\[ f'(t \cdot \frac{1}{t}) \cdot \frac{1}{t} = f'(t) \]Since \( t \cdot \frac{1}{t} = 1 \), this simplifies to:\[ f'(1) \cdot \frac{1}{t} = f'(t) \]Thus, \( f'(t) \) is a function such that \( f'(t) = \frac{f'(1)}{t} \).
3Step 3: Solve differential equation for f(t)
Having determined that \( f'(t) = \frac{f'(1)}{t} \), recognize that this is the derivative of \( f(t) = f'(1) \ln(t) + C \), where \( C \) is a constant of integration.
4Step 4: Evaluate at t = 1 to find C
From Step 3, set \( t = 1 \) in \( f(t) = f'(1) \ln(t) + C \):\[ f(1) = f'(1) \ln(1) + C \]Since \( \ln(1) = 0 \), this simplifies to:\[ f(1) = C \]Therefore, the constant \( C = f(1) \).
5Step 5: Determine f(1) using given property of f
Using the given property \( f(xy) = f(x) + f(y) \) with \( x = 1 \) and \( y = 1 \):\[ f(1 \cdot 1) = f(1) + f(1) \]This results in:\[ f(1) = 2f(1) \]Rearranging gives:\[ f(1) = 0 \]
Key Concepts
AntiderivativesDifferentiationChain RuleLogarithmic Functions
Antiderivatives
Antiderivatives are closely related to differentiation. An antiderivative of a function is essentially the opposite of taking a derivative. If a function is said to be an antiderivative, it means that its derivative returns us to the original function. For example, if we have a function \( f(t) = f'(1) \ln(t) + C \), then its derivative, \( f'(t) \), should give us the function we started with, namely \( \frac{f'(1)}{t} \).
To deduce that \( f(t) = f'(1) \ln(t) + C \) is indeed the antiderivative, we can differentiate \( f(t) \) and confirm it matches \( f'(t) \). This process captures the relationship between a function and its antiderivative. Understanding this gives you the tool to "reverse" differentiation and recover functions from their derivatives.
To deduce that \( f(t) = f'(1) \ln(t) + C \) is indeed the antiderivative, we can differentiate \( f(t) \) and confirm it matches \( f'(t) \). This process captures the relationship between a function and its antiderivative. Understanding this gives you the tool to "reverse" differentiation and recover functions from their derivatives.
Differentiation
Differentiation is the action of computing a derivative. A derivative represents how a function changes as its input changes. In the context of this exercise, differentiation was used to uncover important properties of the function \( f \). By differentiating the given equation \( f(xy) = f(x) + f(y) \) with respect to \( x \), treating \( y \) as a constant, we apply the rules of differentiation.
One key result from differentiation in this exercise is the formula \( f'(xy) \cdot y = f'(x) \). This helps understand how changes in \( x \) relate to changes in \( f(x) \). Differentiation provides a tool for finding rate of change and for solving equations involving calculus by uncovering relationships hidden in the original conditions.
One key result from differentiation in this exercise is the formula \( f'(xy) \cdot y = f'(x) \). This helps understand how changes in \( x \) relate to changes in \( f(x) \). Differentiation provides a tool for finding rate of change and for solving equations involving calculus by uncovering relationships hidden in the original conditions.
Chain Rule
The Chain Rule is a crucial rule in calculus for finding the derivative of a composite function. If a function is composed of multiple inner functions, the chain rule helps us differentiate it efficiently. In this exercise, the chain rule was utilized when differentiating \( f(xy) \) with respect to \( x \), while treating \( y \) as a constant.
By applying the chain rule, we found that the derivative \( \frac{d}{dx}[f(xy)] \) results in \( f'(xy) \cdot y \). This showed how the derivative of the entire function is dependent on both the inner and outer components. Chain Rule simplifies complex differentiation tasks, letting us keep track of how changes in one part of the function affect the whole expression.
By applying the chain rule, we found that the derivative \( \frac{d}{dx}[f(xy)] \) results in \( f'(xy) \cdot y \). This showed how the derivative of the entire function is dependent on both the inner and outer components. Chain Rule simplifies complex differentiation tasks, letting us keep track of how changes in one part of the function affect the whole expression.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions, and they are essential in calculus and various applications. In our specific problem, the function \( f(t) = f'(1) \ln(t) + C \) involves a natural logarithm, \( \ln(t) \).
The logarithmic function in calculus provides us with a way to solve problems where growth and decay rates are proportional to their state, like compound interest or radioactive decay. Understanding their properties, such as \( \ln(1) = 0 \) and how they behave under differentiation \( \frac{d}{dt} [\ln(t)] = \frac{1}{t} \), is essential.
The logarithmic function in calculus provides us with a way to solve problems where growth and decay rates are proportional to their state, like compound interest or radioactive decay. Understanding their properties, such as \( \ln(1) = 0 \) and how they behave under differentiation \( \frac{d}{dt} [\ln(t)] = \frac{1}{t} \), is essential.
- Multiplication inside a logarithm converts to sum: \( \ln(ab) = \ln(a) + \ln(b) \)
- Properties like these proved crucial in linking the given functional equation into a more standard calculus problem.
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