Problem 225 - Calculus Volume 3

Question

Find \(\frac{d f}{d t}\) using the chain rule and direct substitution. \(\quad f(x, y)=\ln (x+y), \quad x=e^{t}, y=e^{t}\)

Step-by-Step Solution

Answer

The derivative \( \frac{df}{dt} = 1 \).

Step 1: Understand the Function

We are given the function \( f(x, y) = \ln(x+y) \) and we are asked to find \( \frac{df}{dt} \), where \( x = e^t \) and \( y = e^t \). This means both \( x \) and \( y \) are functions of \( t \).

Step 2: Apply the Chain Rule

According to the chain rule for functions of two variables, we have:\[\frac{df}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}\]

Step 3: Find Partial Derivatives

First, we need to find the partial derivatives of \( f \): \[\frac{\partial f}{\partial x} = \frac{1}{x+y}\]\[\frac{\partial f}{\partial y} = \frac{1}{x+y}\]

Step 4: Differentiate \( x \) and \( y \) with respect to \( t \)

Since \( x = e^t \) and \( y = e^t \), then:\[\frac{dx}{dt} = e^t\]\[\frac{dy}{dt} = e^t\]

Step 5: Substitute All Values into the Chain Rule Formula

Substitute back into the chain rule equation:\[\frac{df}{dt} = \left( \frac{1}{x+y} \right) e^t + \left( \frac{1}{x+y} \right) e^t \]Simplifying, we get:\[\frac{df}{dt} = \frac{2e^t}{x+y}\]

Step 6: Substitute \( x \) and \( y \)

Recall \( x = e^t \) and \( y = e^t \), so \( x+y = 2e^t \). Substitute this into the equation:\[\frac{df}{dt} = \frac{2e^t}{2e^t} = 1\]

Key Concepts

Partial Derivatives

When dealing with a function of several variables, partial derivatives come into play. They represent how the function changes as you tweak one variable, while keeping others constant. In our exercise, we work with the function \( f(x, y) = \ln(x+y) \). Here, \( x \) and \( y \) are both variables that can alter the function's outcome.
To find the partial derivative of \( f \) with respect to \( x \), we treat \( y \) as a constant. This results in\[\frac{\partial f}{\partial x} = \frac{1}{x+y}\]Similarly, when \( y \) is considered the changing variable and \( x \) is constant:\[\frac{\partial f}{\partial y} = \frac{1}{x+y}\]

• Each partial derivative indicates the rate of change of \( f \) with respect to each variable alone.
• They are key in applying the chain rule where multiple variables depend on another parameter.

Function of Two Variables

Functions of two variables blend two inputs into a single output. Think of it like a machine that needs both \( x \) and \( y \). In our case, \( f(x, y) = \ln(x+y) \), takes fresh \( x \) and \( y \) values and churns out one number by feeding into the logarithmic function. This type of function means:

• For every pair of \( x \) and \( y \), you'll get one output value.
• These functions can represent surfaces plotted in three dimensions, adding a rich layer to analysis.

When we know \( x \) and \( y \) depend on another variable, like \( t \), these combined values drive the result of the function indirectly via the chain rule.

Differentiation with Respect to a Parameter

Differentiation with respect to a parameter such as \( t \) involves tracking how a function's output shifts as that parameter changes. In our problem, both \( x \) and \( y \) vary based on \( t \), specifically \( e^t \). By following the chain rule, we get the change in \( f(x, y) \) as \( t \) shifts:\[\frac{df}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}\]

• The chain rule captures how each variable's change due to \( t \) influences the final result, linking separate effects.
• This approach pieces together changes systematically, vital when calculating rates like speed or growth.

Logarithmic Function Differentiation

Logarithmic functions like \( f(x, y) = \ln(x+y) \) provide an interesting analysis due to their unique properties. When differentiating logs, remember that they follow straightforward rules. For example, differentiating \( \ln(x+y) \) respect to its inputs (\( x \) and \( y \)) simply results in fractions based on the input itself. This simplification:

• Logarithms transform multiplicative relationships into additive ones, simplifying differentiation.
• This results in the uniform output of \( \frac{1}{x+y} \) for each partial derivative, emphasizing their universal rate of change.

Understanding these rules enables you to handle chain rule problems with logarithmic functions and parameters confidently.