Question

find $f_x(x,y)$ and $f_y(x,y)$. then find $f_x(2,-1)$ and $f_y(1,2)$. $f(x,y)=-2e^{6x - 5y}$

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:

\( f_x(x,y)=-12e^{6x - 5y} \), \( f_y(x,y)=10e^{6x - 5y} \), \( f_x(2,-1)=-12e^{17} \), \( f_y(1,2)=\frac{10}{e^{4}} \)