# Solución

Para implementar la regla de la cadena para dos variables, necesitamos seis derivadas parciales: $\frac{\partial z}{\partial x}$, $\frac{\partial z}{\partial y}$, $\frac{\partial x}{\partial x}$, $\frac{\partial x}{\partial v}$, $\frac{\partial y}{\partial u}$ y $\frac{\partial y}{\partial v}$:

$$\frac{\partial z}{\partial x} = 6x-2y$$ | $$\frac{\partial z}{\partial y} = -2x+2y$$
$$\frac{\partial x}{\partial u} = 3$$ | $$\frac{\partial x}{\partial v} = 2$$
$$\frac{\partial y}{\partial u} = 4$$ | $$\frac{\partial y}{\partial v} = -1$$

Para encontrar $\frac{\partial z}{\partial u}$, usamos la ecuación 4.31:

$$\begin{aligned} \frac{\partial z}{\partial u} &= \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}\\ &= 3(6x − 2y) + 4(−2x + 2y)\\ &= 10x+2y \end{aligned}$$

A continuación, sustituimos $x (u, v) = 3u + 2v$ e $y (u, v) = 4u - v$:

$$\begin{aligned} \frac{\partial z}{\partial u} &= 10x+2y\\ &= 10(3u + 2v) + 2(4u − v)\\ &= 38u + 18v \end{aligned}$$

Para encontrar $\frac{\partial z}{\partial v}$, usamos la ecuación 4.32:

$$\begin{aligned} \frac{\partial z}{\partial v} &= \frac{\partial z}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial v}\\ &= 2(6x − 2y) + (−1)(−2x + 2y)\\ &= 14x-6y \end{aligned}$$

Luego sustituimos $x (u, v) = 3u + 2v$ e $y (u, v) = 4u - v$:

$$\begin{aligned} \frac{\partial z}{\partial v} &= 14x-6y\\ &= 14(3u + 2v) − 6(4u − v)\\ &= 18u + 34v \end{aligned}$$