Листик с решением скиньте пжл.

Решите задачу:

$z=sin2xy-cos3xy\; ;\; \; M(\pi ; \frac{1}{6} )\; ;\; \; \vec{l}=(-3;-4)\\\\|\vec{l}|=\sqrt{(-3)^2+(-4)^2}=5\; \; \Rightarrow \quad \vec{l}^\circ=(-\frac{3}{5};-\frac{4}{5})\\\\z'_{x}= 2y\cdot cos2xy+3y\cdot sin3xy\\\\z'_{x}(M)=\frac{1}{3}\cdot cos\frac{\pi}{3}+\frac{1}{2}\cdot sin \frac{\pi }{2}= \frac{1}{3}\cdot \frac{1}{2}+\frac{1}{2}\cdot 1=\frac{1}{6}+\frac{3}{6}=\frac{4}{6}=\frac{2}{3}\\\\z'_{y}=2x\cdot cos2xy+3x\cdot sin3xy\\\\z'_y}(M)=2\pi \cdot cos\frac{\pi}{3}+3\pi \cdot sin\frac{\pi}{2}=2\pi \cdot \frac{1}{2}+3\pi \cdot 1=4\pi$

$\frac{\partial z}{\partial \vec{l}}= \frac{\partial z}{\partial x}\cdot cos \alpha +\frac{\partial z}{\partial y} \cdot cos \beta \\\\\frac{\partial z}{\partial \vec{l}}|_{M} =\frac{\partial z}{\partial x}|_{M}\cdot cos \alpha +\frac{\partial z}{\partial y}|_{M}\cdot cos \beta = \frac{2}{3}\cdot (-\frac{3}{5})+4\pi \cdot (-\frac{4}{5})= \\\\=- \frac{2}{5}-\frac{16\pi }{5}=- \frac{2(1+8\pi )}{5}$