Solución

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

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

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

∂z∂u=∂z∂x∂x∂u+∂z∂y∂y∂u=3(6x−2y)+4(−2x+2y)=10x+2y\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+2vx (u, v) = 3u + 2v e y(u,v)=4u−vy (u, v) = 4u - v:

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

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

∂z∂v=∂z∂x∂x∂v+∂z∂y∂y∂v=2(6x−2y)+(−1)(−2x+2y)=14x−6y\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+2vx (u, v) = 3u + 2v e y(u,v)=4u−vy (u, v) = 4u - v:

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