Question

for the function ( f(x,y)=x^{2}e^{7xy} ), find ( f_{x},f_{y},f_{x}(3,3) ), and ( f_{y}(-3,4) ).

Explanation:

Step1: Find \(f_x\) using product rule

The product rule for differentiation is \((uv)^\prime = u^\prime v+uv^\prime\), where \(u = x^{2}\) and \(v=e^{7xy}\).
\(u^\prime=\frac{d}{dx}(x^{2}) = 2x\), \(v^\prime=\frac{d}{dx}(e^{7xy})=7y e^{7xy}\)
So \(f_x=2x e^{7xy}+x^{2}\cdot7y e^{7xy}=x e^{7xy}(2 + 7xy)\)

Step2: Find \(f_y\) using product rule

Here \(u = x^{2}\) (treated as constant with respect to \(y\)) and \(v = e^{7xy}\)
\(u^\prime=\frac{d}{dy}(x^{2}) = 0\), \(v^\prime=\frac{d}{dy}(e^{7xy})=7x e^{7xy}\)
So \(f_y=x^{2}\cdot7x e^{7xy}=7x^{3}e^{7xy}\)

Step3: Find \(f_x(3,3)\)

Substitute \(x = 3\) and \(y = 3\) into \(f_x\):
\(f_x(3,3)=3e^{7\times3\times3}(2+7\times3\times3)=3e^{63}(2 + 63)=195e^{63}\)

Step4: Find \(f_y(-3,4)\)

Substitute \(x=-3\) and \(y = 4\) into \(f_y\):
\(f_y(-3,4)=7\times(-3)^{3}e^{7\times(-3)\times4}=7\times(-27)e^{-84}=-189e^{-84}\)

Answer:

\(f_x=x e^{7xy}(2 + 7xy)\), \(f_y = 7x^{3}e^{7xy}\), \(f_x(3,3)=195e^{63}\), \(f_y(-3,4)=-189e^{-84}\)