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The problem asks to analyze the function $f(x, y) = e^x \cos y$.

AnalysisPartial DerivativesMultivariable CalculusLaplacianHarmonic Function
2025/4/25

1. Problem Description

The problem asks to analyze the function f(x,y)=excos⁡yf(x, y) = e^x \cos y.

2. Solution Steps

There is no specific instruction. So let's analyze the function f(x,y)=excos⁡yf(x, y) = e^x \cos y.
Partial derivative with respect to x:
∂f∂x=excos⁡y\frac{\partial f}{\partial x} = e^x \cos y
Partial derivative with respect to y:
∂f∂y=−exsin⁡y\frac{\partial f}{\partial y} = -e^x \sin y
Second partial derivative with respect to x:
∂2f∂x2=excos⁡y\frac{\partial^2 f}{\partial x^2} = e^x \cos y
Second partial derivative with respect to y:
∂2f∂y2=−excos⁡y\frac{\partial^2 f}{\partial y^2} = -e^x \cos y
Mixed partial derivative:
∂2f∂x∂y=−exsin⁡y\frac{\partial^2 f}{\partial x \partial y} = -e^x \sin y
∂2f∂y∂x=−exsin⁡y\frac{\partial^2 f}{\partial y \partial x} = -e^x \sin y
Laplacian:
∇2f=∂2f∂x2+∂2f∂y2=excos⁡y−excos⁡y=0\nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = e^x \cos y - e^x \cos y = 0
The function is harmonic.

3. Final Answer

f(x,y)=excos⁡yf(x, y) = e^x \cos y
∂f∂x=excos⁡y\frac{\partial f}{\partial x} = e^x \cos y
∂f∂y=−exsin⁡y\frac{\partial f}{\partial y} = -e^x \sin y
∂2f∂x2=excos⁡y\frac{\partial^2 f}{\partial x^2} = e^x \cos y
∂2f∂y2=−excos⁡y\frac{\partial^2 f}{\partial y^2} = -e^x \cos y
∂2f∂x∂y=−exsin⁡y\frac{\partial^2 f}{\partial x \partial y} = -e^x \sin y
∇2f=0\nabla^2 f = 0

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