Order: z=f(x, y), then:
dz=(? f/? x)dx+(? f/? y)dy
Therefore:
dz=-x? 【e^(-x)]dx+f'y d(x? )={2xf'y-x? [e^(-x)]}dx
At this moment:
z=f(x,x? )=x? [e^(-x]
dz=d{x? [e^(-x)]}=2x[e^(-x)]dx-x? [e^(-x)]dx=(2x-x? )[e^(-x)]dx
∴
2xf'y-x? [e^(-x)]=(2x-x? )[e^(-x)]
f'y=e^(-x)
Choose C.
To answer your question:
1, I don't understand it because you don't understand the role of independent variables, intermediate variables and dependent variables in finding partial derivatives;
2、f'x(x,x? ) is f'x(x, y) in y=x? Instead of f(x, x? ) export values, similar to unary functions, such as:
What does f'(2) mean? Please experience it!
To sum up, your understanding of partial derivatives of binary functions is limited to formulas, but you don't understand them in essence!