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Find the first partial derivatives of the following functions. $$f(x, y)=x e^{y}$$

Short Answer

Expert verified
Question: Find the first partial derivatives of the function \(f(x, y) = xe^{y}\). Answer: The first partial derivatives of the function are: 1. With respect to \(x\): \(\frac{\partial f}{\partial x} = e^{y}\) 2. With respect to \(y\): \(\frac{\partial f}{\partial y} = xe^{y}\).

Step by step solution

01

Find \(\frac{\partial f}{\partial x}\)

First, we compute the partial derivative with respect to the variable \(x\). In doing this, we treat all other variables (in this case, just \(y\)) as constants. Here's the derivative we need to compute: $$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x e^{y})$$ Since \(e^{y}\) is a constant with respect to \(x\), the partial derivative is: $$\frac{\partial f}{\partial x} = e^{y}$$
02

Find \(\frac{\partial f}{\partial y}\)

Now, we compute the partial derivative with respect to the variable \(y\). In doing this, we treat all other variables (in this case, just \(x\)) as constants. Here's the derivative we need to compute: $$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x e^{y})$$ In this case, we use the chain rule for differentiation, which states that if \(h(y) = fg\), then \(h'(y) = f'g + fg'\). Here, \(f = x\) and \(g = e^{y}\). We already know that \(f'(x)=1\) (as computed in Step 1), and \(g'(y) = e^{y}\). Applying the chain rule, we obtain: $$\frac{\partial f}{\partial y} = x \cdot e^{y}$$
03

Final Answer:

The first partial derivatives of the given function, \(f(x, y) = xe^{y}\), are as follows: 1. The partial derivative of \(f\) with respect to \(x\): \(\frac{\partial f}{\partial x} = e^{y}\). 2. The partial derivative of \(f\) with respect to \(y\): \(\frac{\partial f}{\partial y} = xe^{y}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The Chain Rule is a fundamental concept in calculus used to find the derivative of composite functions. It allows us to differentiate a function based on its internal structure, applying the derivative rule to each part of the function. This is crucial when dealing with multivariable functions, like in our original exercise.

In the context of our function, \[f(x,y) = xe^{y},\] the chain rule helps us find the partial derivative with respect to a certain variable by treating other variables as constants. For example, when computing \(\frac{\partial f}{\partial y}\), we treat \(x\) as constant. We can think of \(x\) as one function \((f)\) and \(e^y\) as another \((g)\). Using the product rule in combination with the chain rule, we calculate the derivative by differentiating each part:

\[h'(y) = f'g + fg'.\] Here, \(f' = 0\) since \(x\) is a constant with respect to \(y\), and \(g' = e^{y}\), leading us to:

\[\frac{\partial f}{\partial y} = xe^{y}.\] Chains of derivatives are connected step-by-step, making calculus a seamless process of unraveling complexities in functions.
Multivariable Calculus
Multivariable Calculus expands the concept of calculus to functions with more than one variable. It's like regular calculus but instead of moving in a straight line, it explores surfaces and shapes, with those extra dimensions adding depth and complexity.

In this realm, we're interested in how changing one variable affects the outcome while the others remain constant, which is where partial derivatives come in. Take our function \(f(x, y) = xe^{y}\). To understand its behavior completely, we compute the partial derivatives with respect to each variable separately; one with \(x\) (holding \(y\) constant) and one with \(y\) (holding \(x\) constant).

Understanding these concepts is vital for fields like engineering, physics, and economics because they involve systems that change in multiple dimensions. The way you navigate and predict these systems often relies heavily on knowing how each variable independently influences the result.
Differentiation
Differentiation is the process of finding the rate at which a function is changing at any given point and is one of the core operations in calculus. When applying differentiation in the context of multivariable functions, we focus on partial derivatives, which show us how a change in one variable affects the function’s output, with other variables fixed.

The original exercise dealt with the differentiation of the function \(f(x,y) = xe^y\). For the partial derivative with respect to \(x\), \(\frac{\partial f}{\partial x} = e^y\), we examine how \(f\) changes as \(x\) changes, treating \(y\) as a fixed constant.

For the partial derivative with respect to \(y\), we apply the chain rule to see how \(f\) changes with \(y\), holding \(x\) fixed. This results in \(\frac{\partial f}{\partial y} = xe^y\).

These steps not only deepen our understanding of the function's sensitivity to changes in its variables but also highlight the methodical thinking required when tackling complex functions involving several variables. Differentiation in multivariable settings provides crucial insight into the dynamics of functions across multidimensional spaces.

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