find $f_{x}(x,y)$ and $f_{y}(x,y)$. then find $f_{x}(2,-1)$ and $f_{y}(-4,3)$. $f(x,y)=e^{x + y + 3}$…

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.)

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.)

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})