Problem 12 - Calculus Early Transcendentals

Question

Find the first partial derivatives of the following functions. $$f(x, y)=\ln (x / y)$$

Step-by-Step Solution

Answer

Answer: The first partial derivatives are \(\frac{\partial f}{\partial x} = \frac{1}{x}\) and \(\frac{\partial f}{\partial y} = -\frac{1}{y}\).

Step 1: Rewrite the function using logarithm properties

We can rewrite the function using the property of logarithms: \(\ln(a / b) = \ln(a) - \ln(b)\). So, we can rewrite the given function as $$f(x, y) = \ln(x) - \ln(y).$$

Step 2: Find the partial derivative with respect to x

We can now find the partial derivative of \(f(x, y)\) with respect to \(x\), denoted as \(\frac{\partial f}{\partial x}\). Using the properties of logarithm, we have: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(\ln(x) - \ln(y)).$$ Since \(\ln(y)\) is a constant with respect to \(x\), its derivative will be zero. Therefore: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(\ln(x)) - 0$$ The derivative of \(\ln(x)\) with respect to \(x\) is \(\frac{1}{x}\). So, we have: $$\frac{\partial f}{\partial x} = \frac{1}{x}.$$

Step 3: Find the partial derivative with respect to y

Now, we find the partial derivative of \(f(x, y)\) with respect to \(y\), denoted as \(\frac{\partial f}{\partial y}\). Using the properties of logarithm, we have: $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(\ln(x) - \ln(y)).$$ Since \(\ln(x)\) is a constant with respect to \(y\), its derivative will be zero. Therefore: $$\frac{\partial f}{\partial y} = 0 - \frac{\partial}{\partial y}(\ln(y))$$ The derivative of \(\ln(y)\) with respect to \(y\) is \(\frac{1}{y}\). So, we have: $$\frac{\partial f}{\partial y} = -\frac{1}{y}.$$

Step 4: Write down the first partial derivatives

We've found the first partial derivatives of the given function, \(f(x, y) = \ln(x / y)\). The first partial derivatives are: $$\frac{\partial f}{\partial x} = \frac{1}{x}$$ and $$\frac{\partial f}{\partial y} = -\frac{1}{y}.$$