find $f_x(x,y)$ and $f_y(x,y)$. then find $f_x(2,-1)$ and $f_y(1,2)$. $f(x,y)=-2e^{6x - 5y}$
find $f_x(x,y)$ and $f_y(x,y)$. then find $f_x(2,-1)$ and $f_y(1,2)$. $f(x,y)=-2e^{6x - 5y}$
Answer
Explanation:
Step1: Find ( f_x(x,y) )
Differentiate ( f(x,y)=-2e^{6x - 5y} ) with respect to ( x ) using the chain rule. If ( u = 6x-5y ), then ( \frac{\partial f}{\partial x}=-2e^{u}\times\frac{\partial u}{\partial x} ). Since ( \frac{\partial u}{\partial x}=6 ), we have ( f_x(x,y)=-12e^{6x - 5y} ).
Step2: Find ( f_y(x,y) )
Differentiate ( f(x,y)=-2e^{6x - 5y} ) with respect to ( y ) using the chain rule. If ( u = 6x-5y ), then ( \frac{\partial f}{\partial y}=-2e^{u}\times\frac{\partial u}{\partial y} ). Since ( \frac{\partial u}{\partial y}=-5 ), we have ( f_y(x,y)=10e^{6x - 5y} ).
Step3: Find ( f_x(2,-1) )
Substitute ( x = 2 ) and ( y=-1 ) into ( f_x(x,y) ). ( f_x(2,-1)=-12e^{6\times2-5\times(-1)}=-12e^{12 + 5}=-12e^{17} ).
Step4: Find ( f_y(1,2) )
Substitute ( x = 1 ) and ( y = 2 ) into ( f_y(x,y) ). ( f_y(1,2)=10e^{6\times1-5\times2}=10e^{6-10}=10e^{-4}=\frac{10}{e^{4}} ).
Answer:
( f_x(x,y)=-12e^{6x - 5y} ), ( f_y(x,y)=10e^{6x - 5y} ), ( f_x(2,-1)=-12e^{17} ), ( f_y(1,2)=\frac{10}{e^{4}} )