Question:

Step 1: Simplify the functional equation The given functional equation is $2f(x) = f(x) + f\left(\dfrac{x}{y}\right)$ for all $x, y > 0$. Subtracting $f(x)$ from both sides yields: $$f(x) = f\left(\dfrac{x}{y}\right)$$ Step 2: Determine the derivative of $f(x)$ Differentiate the simplified functional equation $f(x) = f\left(\dfrac{x}{y}\right)$ with respect to $x$. Applying the chain rule to the right side: $$f'(x) = f'\left(\dfrac{x}{y}\right) \cdot \dfrac{\partial}{\partial x}\left(\dfrac{x}{y}\right)$$ $$f'(x) = f'\left(\dfrac{x}{y}\right) \cdot \dfrac{1}{y}$$ Substitute $x=1$ into this equation: $$f'(1) = f'\left(\dfrac{1}{y}\right) \cdot \dfrac{1}{y}$$ Given $f'(1) = 1$: $$1 = f'\left(\dfrac{1}{y}\right) \cdot \dfrac{1}{y}$$ Multiplying by $y$ gives: $$f'\left(\dfrac{1}{y}\right) = y$$ Let $t = \dfrac{1}{y}$. As $y$ ranges over $(0, \infty)$, $t$ also ranges over $(0, \infty)$. Thus, $f'(t) = \dfrac{1}{t}$ for all $t > 0$. Step 3: Solve for $f(x)$ Integrate $f'(t) = \dfrac{1}{t}$ with respect to $t$: $$f(t) = \int \dfrac{1}{t} \, dt = \ln|t| + C$$ Since the domain of $f$ is $(0, \infty)$, we can write $f(t) = \ln(t) + C$. Using the given condition $f(1) = 0$: $$f(1) = \ln(1) + C = 0 + C = 0$$ Therefore, $C = 0$. The function is $f(x) = \ln(x)$. Step 4: Verify initial conditions The derived function $f(x) = \ln(x)$ satisfies the given initial conditions: 1. $f(1) = \ln(1) = 0$. 2. $f'(x) = \dfrac{1}{x}$, so $f'(1) = \dfrac{1}{1} = 1$.