Math Problem Statement

Find the partial derivatives of the function f(x, y) = xye^{1/y}.

Solution

To solve this problem, we need to find the partial derivatives fx(x,y)f_x(x, y), fy(x,y)f_y(x, y), fxy(x,y)f_{xy}(x, y), and fyx(x,y)f_{yx}(x, y) of the function:

f(x,y)=xye1/yf(x, y) = x y e^{1/y}

Step 1: Calculate fx(x,y)f_x(x, y)

The partial derivative of ff with respect to xx is found by differentiating f(x,y)=xye1/yf(x, y) = x y e^{1/y} with respect to xx, treating yy as a constant.

fx(x,y)=x(xye1/y)=ye1/yf_x(x, y) = \frac{\partial}{\partial x} \left( x y e^{1/y} \right) = y e^{1/y}

Step 2: Calculate fy(x,y)f_y(x, y)

The partial derivative of ff with respect to yy is found by differentiating f(x,y)=xye1/yf(x, y) = x y e^{1/y} with respect to yy, treating xx as a constant.

To differentiate with respect to yy, we apply the product rule, as f(x,y)=x(ye1/y)f(x, y) = x \cdot (y e^{1/y}):

fy(x,y)=xddy(ye1/y)f_y(x, y) = x \cdot \frac{d}{dy}\left(y e^{1/y}\right)

Now, differentiate ye1/yy e^{1/y} with respect to yy:

ddy(ye1/y)=e1/y+yddy(e1/y)\frac{d}{dy}\left(y e^{1/y}\right) = e^{1/y} + y \cdot \frac{d}{dy}\left(e^{1/y}\right)

Since ddy(e1/y)=e1/y(1y2)\frac{d}{dy}\left(e^{1/y}\right) = e^{1/y} \cdot \left(-\frac{1}{y^2}\right):

=e1/ye1/yy=e1/y(11y)= e^{1/y} - \frac{e^{1/y}}{y} = e^{1/y} \left(1 - \frac{1}{y}\right)

Thus,

fy(x,y)=xe1/y(11y)=xe1/y(11y)f_y(x, y) = x \cdot e^{1/y} \left(1 - \frac{1}{y}\right) = x e^{1/y} \left(1 - \frac{1}{y}\right)

Step 3: Calculate fxy(x,y)f_{xy}(x, y)

The mixed partial derivative fxy(x,y)f_{xy}(x, y) is the partial derivative of fx(x,y)=ye1/yf_x(x, y) = y e^{1/y} with respect to yy.

fxy(x,y)=y(ye1/y)f_{xy}(x, y) = \frac{\partial}{\partial y} \left(y e^{1/y}\right)

As computed earlier:

fxy(x,y)=e1/y(11y)f_{xy}(x, y) = e^{1/y} \left(1 - \frac{1}{y}\right)

Step 4: Calculate fyx(x,y)f_{yx}(x, y)

Since mixed partial derivatives are usually equal if the function is continuous and well-behaved, we have:

fyx(x,y)=fxy(x,y)=e1/y(11y)f_{yx}(x, y) = f_{xy}(x, y) = e^{1/y} \left(1 - \frac{1}{y}\right)

Summary of Results

  1. fx(x,y)=ye1/yf_x(x, y) = y e^{1/y}
  2. fy(x,y)=xe1/y(11y)f_y(x, y) = x e^{1/y} \left(1 - \frac{1}{y}\right)
  3. fxy(x,y)=e1/y(11y)f_{xy}(x, y) = e^{1/y} \left(1 - \frac{1}{y}\right)
  4. fyx(x,y)=e1/y(11y)f_{yx}(x, y) = e^{1/y} \left(1 - \frac{1}{y}\right)

Would you like further details or have any questions?


Related Questions:

  1. How do we determine if mixed partial derivatives are equal?
  2. What is the significance of the exponential function e1/ye^{1/y} in partial derivatives?
  3. How does the product rule apply in partial differentiation?
  4. What happens if y=0y = 0 in this function?
  5. Can you apply these derivatives to analyze the behavior of f(x,y)f(x, y) near specific points?

Tip: Always check for continuity when assuming mixed partial derivatives are equal.

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Math Problem Analysis

Mathematical Concepts

Calculus
Partial Derivatives
Exponential Functions

Formulas

f_x(x, y) = ∂f/∂x
f_y(x, y) = ∂f/∂y
f_{xy}(x, y) = ∂²f/∂x∂y
f_{yx}(x, y) = ∂²f/∂y∂x

Theorems

Mixed Derivative Theorem

Suitable Grade Level

Undergraduate Calculus