# Solución

Las fórmulas para $\frac{\partial w}{\partial u}$ y $\frac{\partial w}{\partial v}$ son

$$\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial u}$$ $$\frac{\partial w}{\partial v} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial v} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial v}$$

Por lo tanto, hay nueve derivadas parciales diferentes que deben calcularse y sustituirse. Necesitamos calcular cada una de ellas:

$$\frac{\partial w}{\partial x} = 6x-2y$$ | $$\frac{\partial w}{\partial y} = -2x$$ | $$\frac{\partial w}{\partial z} = 8z$$
$$\frac{\partial x}{\partial u} = e^usenv$$ | $$\frac{\partial y}{\partial u} = e^ucosv$$ | $$\frac{\partial z}{\partial u} = e^u$$
$$\frac{\partial x}{\partial v} = e^ucosv$$ | $$\frac{\partial y}{\partial v} = -e^usenv$$ | $$\frac{\partial z}{\partial v} = 0$$

Ahora, sustituimos cada una de ellas en la primera fórmula para calcular $\frac{\partial w}{\partial u}$:

$$\begin{aligned} \frac{\partial w}{\partial u} &= \frac{\partial w}{\partial x}\cdot\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\cdot\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\cdot\frac{\partial z}{\partial u}\\ &= (6x − 2y)e^usen v − 2xe^ucos v + 8ze^u \end{aligned}$$

entonces sustituye $x (u, v) = e^usen v$, $y (u, v) = e^ucos v$, y $z (u, v) = e^u$ en esta ecuación:

$$\begin{aligned} \frac{\partial w}{\partial u} &= (6x − 2y)e^usen v − 2xe^ucos v + 8ze^u\\ &= (6e^usen v − 2e^ucos v)e^usen v − 2(e^usen v)e^ucos v + 8e^{2u}\\ &= 6e^{2u}sen^2v − 4e^{2u}sen v cos v + 8e^{2u}\\ &= 2e^{2u}\big(3 sen^2v − 2 sin v cos v + 4\big) \end{aligned}$$

A continuación, calculamos $\frac{\partial w}{\partial v}$:

$$\begin{aligned} \frac{\partial w}{\partial v} &= \frac{\partial w}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial v} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial v}\\ &= (6x − 2y)e^ucos v − 2x(−e^usen v) + 8z(0) \end{aligned}$$

entonces sustituimos $x (u, v) = e^usen v$, $y (u, v) = e^ucos v$, y $z (u, v) = e^u$ en esta ecuación:

$$\begin{aligned} \frac{\partial w}{\partial v} &= (6x − 2y)e^ucos v − 2x(−e^usen v)\\ &= (6e^usen v − 2e^ucos v)e^ucos v + 2(e^usen v)(e^usen v)\\ &= 2e^{2u}sen^2v + 6e^{2u}sen v cos v − 2e^{2u}cos^2v\\ &= 2e^{2u}\big(sen^2v + sen v cos v − cos^2v\big) \end{aligned}$$