Question

1. find the $f_x(x,y)$, $f_y(x,y)$, $f_x(1,2)$, and $f_y(1,2)$

$f(x,y)=xcdot e^{x^2y}$

Explanation:

Step1: Find \(f_x(x,y)\)

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x\), \(v=e^{x^{2}y}\).
The derivative of \(u=x\) with respect to \(x\) is \(u^\prime = 1\).
The derivative of \(v = e^{x^{2}y}\) with respect to \(x\) is \(v^\prime=e^{x^{2}y}\cdot2xy\) (by the chain rule \((e^{g(x)})^\prime=e^{g(x)}\cdot g^\prime(x)\), here \(g(x)=x^{2}y\)).
So \(f_x(x,y)=e^{x^{2}y}+x\cdot e^{x^{2}y}\cdot2xy=e^{x^{2}y}(1 + 2x^{2}y)\).

Step2: Find \(f_y(x,y)\)

Treat \(x\) as a constant. Use the chain rule.
If \(f(x,y)=x\cdot e^{x^{2}y}\), then \(f_y(x,y)=x\cdot e^{x^{2}y}\cdot x^{2}=x^{3}e^{x^{2}y}\).

Step3: Find \(f_x(1,2)\)

Substitute \(x = 1\) and \(y = 2\) into \(f_x(x,y)\).
\(f_x(1,2)=e^{1^{2}\times2}(1+2\times1^{2}\times2)=e^{2}(1 + 4)=5e^{2}\).

Step4: Find \(f_y(1,2)\)

Substitute \(x = 1\) and \(y = 2\) into \(f_y(x,y)\).
\(f_y(1,2)=1^{3}\cdot e^{1^{2}\times2}=e^{2}\).

Answer:

\(f_x(x,y)=e^{x^{2}y}(1 + 2x^{2}y)\), \(f_y(x,y)=x^{3}e^{x^{2}y}\), \(f_x(1,2)=5e^{2}\), \(f_y(1,2)=e^{2}\)