Question

find $f_{x}(x,y)$ and $f_{y}(x,y)$. then find $f_{x}(2,-1)$ and $f_{y}(-4,3)$. $f(x,y)=e^{x + y + 3}$ $f_{x}(x,y)=\square$ $f_{y}(x,y)=\square$ $f_{x}(2,-1)=\square$ (type an exact answer.) $f_{y}(-4,3)=\square$ (type an exact answer.)

Explanation:

Step1: Find partial derivatives

For \(f(x,y)=e^{x + y+3}\), use the chain - rule. The partial derivative of \(e^{u}\) with respect to \(x\) (where \(u=x + y+3\)) is \(e^{u}\cdot\frac{\partial u}{\partial x}\), and the partial derivative of \(e^{u}\) with respect to \(y\) (where \(u=x + y+3\)) is \(e^{u}\cdot\frac{\partial u}{\partial y}\).
Since \(\frac{\partial(x + y+3)}{\partial x}=1\) and \(\frac{\partial(x + y+3)}{\partial y}=1\), we have \(f_{x}(x,y)=e^{x + y+3}\) and \(f_{y}(x,y)=e^{x + y+3}\).

Step2: Evaluate \(f_{x}(2,-1)\)

Substitute \(x = 2\) and \(y=-1\) into \(f_{x}(x,y)\).
\(f_{x}(2,-1)=e^{2+( - 1)+3}=e^{4}\).

Step3: Evaluate \(f_{y}(-4,3)\)

Substitute \(x=-4\) and \(y = 3\) into \(f_{y}(x,y)\).
\(f_{y}(-4,3)=e^{-4 + 3+3}=e^{2}\).

Answer:

\(f_{x}(x,y)=e^{x + y+3}\), \(f_{y}(x,y)=e^{x + y+3}\), \(f_{x}(2,-1)=e^{4}\), \(f_{y}(-4,3)=e^{2}\)