To solve this problem, we need to find the partial derivatives fx(x,y), fy(x,y), fxy(x,y), and fyx(x,y) of the function:
f(x,y)=xye1/y
Step 1: Calculate fx(x,y)
The partial derivative of f with respect to x is found by differentiating f(x,y)=xye1/y with respect to x, treating y as a constant.
fx(x,y)=∂x∂(xye1/y)=ye1/y
Step 2: Calculate fy(x,y)
The partial derivative of f with respect to y is found by differentiating f(x,y)=xye1/y with respect to y, treating x as a constant.
To differentiate with respect to y, we apply the product rule, as f(x,y)=x⋅(ye1/y):
fy(x,y)=x⋅dyd(ye1/y)
Now, differentiate ye1/y with respect to y:
dyd(ye1/y)=e1/y+y⋅dyd(e1/y)
Since dyd(e1/y)=e1/y⋅(−y21):
=e1/y−ye1/y=e1/y(1−y1)
Thus,
fy(x,y)=x⋅e1/y(1−y1)=xe1/y(1−y1)
Step 3: Calculate fxy(x,y)
The mixed partial derivative fxy(x,y) is the partial derivative of fx(x,y)=ye1/y with respect to y.
fxy(x,y)=∂y∂(ye1/y)
As computed earlier:
fxy(x,y)=e1/y(1−y1)
Step 4: Calculate fyx(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(1−y1)
Summary of Results
- fx(x,y)=ye1/y
- fy(x,y)=xe1/y(1−y1)
- fxy(x,y)=e1/y(1−y1)
- fyx(x,y)=e1/y(1−y1)
Would you like further details or have any questions?
Related Questions:
- How do we determine if mixed partial derivatives are equal?
- What is the significance of the exponential function e1/y in partial derivatives?
- How does the product rule apply in partial differentiation?
- What happens if y=0 in this function?
- Can you apply these derivatives to analyze the behavior of f(x,y) near specific points?
Tip: Always check for continuity when assuming mixed partial derivatives are equal.