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

Question

Accepted Answer

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

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QUESTION IMAGE

Question

Explanation:

Step1: Find \(f_x\)

Step2: Evaluate \(f_x(3,2)\)

Step3: Find \(f_y\)

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

Answer: