find the differential of the function $f(x,y)=ye^{-x}$ at $(0,2)$. $df=$

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 ((uv)^\prime = u^\prime v+uv^\prime) (here (u = y), (v = e^{-x})), (\frac{\partial f}{\partial x}=y\frac{\partial}{\partial x}(e^{-x})=-ye^{-x}). Find (\frac{\partial f}{\partial y}): (\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}(ye^{-x})=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}\big|{(0,2)}=-2e^{0}=- 2). Substitute (x = 0) into (\frac{\partial f}{\partial y}): (\frac{\partial f}{\partial y}\big|{(0,2)}=e^{0}=1).

Step3: Write the differential

Since (df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy), substituting the values of (\frac{\partial f}{\partial x}) and (\frac{\partial f}{\partial y}) at ((0,2)) gives (df=-2dx + 1dy).

Answer:

(df=-2dx+dy)