for the function ( f(x,y)=x^{2}e^{5xy} ), find ( f_{x},f_{y},f_{x}(3,2) ), and ( f_{y}(-3,-4) ).\n(…

Explanation:

Step1: Find (f_x)

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{2}), (u^\prime=2x), (v = e^{5xy}), (v^\prime=5ye^{5xy}). [ \begin{align*} f_x&=(x^{2})^\prime e^{5xy}+x^{2}(e^{5xy})^\prime\ &=2x e^{5xy}+x^{2}\cdot5y e^{5xy}\ &=5x^{2}ye^{5xy}+2xe^{5xy} \end{align*} ]

Step2: Evaluate (f_x(3,2))

Substitute (x = 3) and (y = 2) into (f_x). [ \begin{align*} f_x(3,2)&=5\times3^{2}\times2\times e^{5\times3\times2}+2\times3\times e^{5\times3\times2}\ &=90e^{30}+6e^{30}\ &=(90 + 6)e^{30}\ &=96e^{30} \end{align*} ]

Step3: Find (f_y)

Use the chain - rule. Let (u = 5xy), then (\frac{\partial f}{\partial y}=x^{2}\cdot5x e^{5xy}=5x^{3}e^{5xy})

Step4: Evaluate (f_y(-3,-4))

Substitute (x=-3) and (y = - 4) into (f_y). [ \begin{align*} f_y(-3,-4)&=5\times(-3)^{3}\times e^{5\times(-3)\times(-4)}\ &=5\times(-27)\times e^{60}\ &=-135e^{60} \end{align*} ]

Answer:

(f_x = 5x^{2}ye^{5xy}+2xe^{5xy}), (f_y=5x^{3}e^{5xy}), (f_x(3,2)=96e^{30}), (f_y(-3,-4)=-135e^{60})