find the partial derivatives of the function (f(x,y)=\frac{-7x + 5y}{-7x+8y}). (f_x(x,y)=) (f_y(x,y)=)

Explanation:

Step1: Recall partial - derivative formula

For a function $z = \frac{u}{v}$, the partial - derivative with respect to $x$ (treating $y$ as a constant) is given by $\frac{\partial z}{\partial x}=\frac{u_xv - uv_x}{v^2}$, and with respect to $y$ (treating $x$ as a constant) is given by $\frac{\partial z}{\partial y}=\frac{u_yv - uv_y}{v^2}$, where $u_x=\frac{\partial u}{\partial x}$, $u_y=\frac{\partial u}{\partial y}$, $v_x=\frac{\partial v}{\partial x}$, and $v_y=\frac{\partial v}{\partial y}$. Here, $u=-7x + 5y$, $v=-7x + 8y$.

Step2: Find $f_x(x,y)$

First, find the partial - derivatives of $u$ and $v$ with respect to $x$: $u_x=-7$, $v_x=-7$. Then, by the quotient rule for partial derivatives $\frac{\partial f}{\partial x}=\frac{u_xv - uv_x}{v^2}=\frac{(-7)(-7x + 8y)-(-7x + 5y)(-7)}{(-7x + 8y)^2}=\frac{49x-56y-(49x - 35y)}{(-7x + 8y)^2}=\frac{49x-56y - 49x + 35y}{(-7x + 8y)^2}=\frac{-21y}{(-7x + 8y)^2}$.

Step3: Find $f_y(x,y)$

Find the partial - derivatives of $u$ and $v$ with respect to $y$: $u_y = 5$, $v_y=8$. Then, by the quotient rule for partial derivatives $\frac{\partial f}{\partial y}=\frac{u_yv - uv_y}{v^2}=\frac{5(-7x + 8y)-(-7x + 5y)\times8}{(-7x + 8y)^2}=\frac{-35x + 40y-(-56x + 40y)}{(-7x + 8y)^2}=\frac{-35x + 40y + 56x-40y}{(-7x + 8y)^2}=\frac{21x}{(-7x + 8y)^2}$.

Answer:

$f_x(x,y)=\frac{-21y}{(-7x + 8y)^2}$, $f_y(x,y)=\frac{21x}{(-7x + 8y)^2}$