Question

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:

\(df=-2dx+dy\)