Question

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

\(df=-2dx+dy\)