# Solución

Primero, calcula $f_x(x, y)$ y $f_y(x, y)$, luego usa la ecuación 4.24 con $x_0 = \pi/3$ y $y_0 = \pi/4$:

$$\begin{aligned} f_x(x, y) &= 2cos(2x)cos(3y)\\ f_y(x, y) &= −3sen(2x)sen(3y)\\ f\bigg(\frac{\pi}{3}, \frac{\pi}{4}\bigg) &= sen\bigg(2\bigg(\frac{\pi}{3}\bigg)\bigg)cos\bigg(3\bigg(\frac{\pi}{4}\bigg)\bigg) = \bigg(\frac{\sqrt{3}}{2}\bigg)\bigg(-\frac{\sqrt{2}}{2}\bigg) = -\frac{\sqrt{6}}{4}\\ f_x\bigg(\frac{\pi}{3}, \frac{\pi}{4}\bigg) &= 2cos\bigg(2\bigg(\frac{\pi}{3}\bigg)\bigg)cos\bigg(3\bigg(\frac{\pi}{4}\bigg)\bigg) = 2\bigg(-\frac12\bigg)\bigg(-\frac{\sqrt{2}}{2}\bigg) = \frac{\sqrt{2}}{2}\\ f_y\bigg(\frac{\pi}{3}, \frac{\pi}{4}\bigg) &= -3sen\bigg(2\bigg(\frac{\pi}{3}\bigg)\bigg)sen\bigg(3\bigg(\frac{\pi}{4}\bigg)\bigg) = -3\bigg(\frac{\sqrt{3}}{2}\bigg)\bigg(\frac{\sqrt{2}}{2}\bigg) = -\frac{3\sqrt{6}}{4} \end{aligned}$$

Entonces la ecuación 4.24 se convierte en

$$\begin{aligned} z &= f(x_0, y_0) + f_x(x_0, y_0)(x − x_0) + f_y(x_0, y_0)(y − y_0)\\ z &= -\frac{\sqrt{6}}{5}+\frac{\sqrt{2}}{2}\bigg(x-\frac{\pi}{3}\bigg) - \frac{3\sqrt{6}}{4}\bigg(y-\frac{\pi}{4}\bigg)\\ x &= \frac{\sqrt{2}}{2}x - \frac{3\sqrt{6}}{4}y - \frac{\sqrt{6}}{4} - \frac{\pi\sqrt{2}}{6} + \frac{3\pi\sqrt{6}}{16} \end{aligned}$$