find the differential of the function ( f(x, y)=y e^{-x} ) at ( (0,2) ).\n( d f= )

Explanation:

Step1: Find partial derivatives

The formula for the differential (df) of a function (z = f(x,y)) is (df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy). For (f(x,y)=ye^{-x}), find (\frac{\partial f}{\partial x}) using the product rule. (\frac{\partial f}{\partial x}=-ye^{-x}). Find (\frac{\partial f}{\partial y}) using the power rule. (\frac{\partial f}{\partial y}=e^{-x}).

Step2: Evaluate partial derivatives at ((0,2))

Substitute (x = 0) and (y = 2) into (\frac{\partial f}{\partial x}): (\frac{\partial f}{\partial x}\mid_{(0,2)}=-2e^{0}=-2). Substitute (x = 0) into (\frac{\partial f}{\partial y}): (\frac{\partial f}{\partial y}\mid_{(0,2)}=e^{0}=1).

Step3: Write the differential

Substitute (\frac{\partial f}{\partial x}\mid_{(0,2)}=-2) and (\frac{\partial f}{\partial y}\mid_{(0,2)} = 1) into (df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy). We get (df=-2dx + 1dy).

Answer:

(df=-2dx+dy)