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})