find $f_x(x,y)$ and $f_y(x,y)$. then, find $f_x(2, - 1)$ and $f_y(-4, - 2)$. $f(x,y)=ln|8 + 3x^{2}y^{3}|$…

Explanation:

Step1: Use chain - rule for partial derivative w.r.t x

The derivative of $\ln(u)$ is $\frac{u'}{u}$. Here $u = 8 + 3x^{2}y^{3}$. Differentiating $u$ with respect to $x$ gives $\frac{\partial u}{\partial x}=6xy^{3}$. So, $f_x(x,y)=\frac{6xy^{3}}{8 + 3x^{2}y^{3}}$.

Step2: Use chain - rule for partial derivative w.r.t y

Differentiating $u = 8+3x^{2}y^{3}$ with respect to $y$ gives $\frac{\partial u}{\partial y}=9x^{2}y^{2}$. So, $f_y(x,y)=\frac{9x^{2}y^{2}}{8 + 3x^{2}y^{3}}$.

Step3: Evaluate $f_x(2,-1)$

Substitute $x = 2$ and $y=-1$ into $f_x(x,y)$: [ \begin{align*} f_x(2,-1)&=\frac{6\times2\times(-1)^{3}}{8 + 3\times2^{2}\times(-1)^{3}}\ &=\frac{12\times(-1)}{8+12\times(-1)}\ &=\frac{-12}{8 - 12}\ &=\frac{-12}{-4}\ & = 3 \end{align*} ]

Step4: Evaluate $f_y(-4,-2)$

Substitute $x=-4$ and $y = - 2$ into $f_y(x,y)$: [ \begin{align*} u&=8+3\times(-4)^{2}\times(-2)^{3}\ &=8+3\times16\times(-8)\ &=8 - 384\ &=-376 \end{align*} ] [ \begin{align*} f_y(-4,-2)&=\frac{9\times(-4)^{2}\times(-2)^{2}}{8 + 3\times(-4)^{2}\times(-2)^{3}}\ &=\frac{9\times16\times4}{-376}\ &=\frac{576}{-376}\ &=-\frac{72}{47} \end{align*} ]

Answer:

$f_x(x,y)=\frac{6xy^{3}}{8 + 3x^{2}y^{3}}$, $f_y(x,y)=\frac{9x^{2}y^{2}}{8 + 3x^{2}y^{3}}$, $f_x(2,-1)=3$, $f_y(-4,-2)=-\frac{72}{47}$