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Multivariable calculus is an extension of calculus to functions with more than one variable. It provides powerful tools for understanding how changes in each variable affect the function as a whole. When dealing with functions of several variables, such as two variables like in the problem presented, we often need to determine how the function behaves as we vary one of the variables while keeping the other constant. This involves finding partial derivatives.

In this exercise, you work with a function, \(f(x, y) = \ln(x + y)\). Here, the function depends both on \(x\) and \(y\). Our job is to determine the effect of modifying \(x\) or \(y\) independently by finding \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).

• Partial derivatives measure how a function changes as each variable alone is increased by a small increment.
• The derivative \(\frac{\partial f}{\partial x} = \frac{1}{x+y}\) reveals how the function changes with \(x\) when \(y\) stays constant.
• Similarly, \(\frac{\partial f}{\partial y} = \frac{1}{x+y}\) tells us the change with \(y\) when \(x\) is unchanged.

In practice, these calculations are crucial in various fields such as physics, engineering, and economics, where models often involve many variables interacting simultaneously.

Logarithmic differentiation is a technique used to differentiate functions that involve logarithms, allowing us to handle complex expressions more easily. In our problem, the function \(f(x, y) = \ln(x + y)\) naturally lends itself to this method because it includes a natural logarithm.

The principle behind logarithmic differentiation is that if you have \(\ln(u(x))\), where \(u(x)\) is some function of \(x\), the derivative is \(\frac{1}{u(x)} \cdot u'(x)\). For a multivariable function, this principle extends in a similar manner. Here, when finding \(\frac{\partial f}{\partial x}\), treat \(y\) as a constant such that the derivative of \(u = x + y\) with respect to \(x\) is \(1\). Thus:

• \(\frac{\partial f}{\partial x} = \frac{1}{x+y} \cdot 1 = \frac{1}{x+y}\)

Likewise, for \(\frac{\partial f}{\partial y}\), treat \(x\) as a constant and recognize the similarity in calculation:

• \(\frac{\partial f}{\partial y} = \frac{1}{x+y} \cdot 1 = \frac{1}{x+y}\)

This process allows for simplified computation even if the form of the function appears initially daunting.

A function of two variables, like \(f(x, y) = \ln(x+y)\), maps pairs of values from two different inputs to a single output. Understanding functions of two variables requires analyzing how changes in each variable affect the output.

For any given point in the \(xy\)-plane, the value of \(f\) depends on both \(x\) and \(y\). To study these dependencies, partial derivatives are a key tool, as they tell us how the function behaves when only one input is varied at a time. This is crucial for graphing and modeling the function's surface:

• Imagine a landscape where every point \((x, y)\) has an elevation corresponding to \(f(x, y)\).
• Partial derivatives indicate the steepness of the slope in the \(x\)-direction and the \(y\)-direction.

By calculating \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) for all \((x, y)\), we can understand how sharp or gentle the landscape is across different directions.

This approach not only helps in visualizing the function but also plays an essential role in optimization and constraint problems, where we seek to maximize or minimize such functions.