Даны: Функция u=arctg(x*y)+z^3, точка B (1,-1,2) и вектор а=-2i+3j+. Найти gradu в точке...

Решите задачу:

$u=\arctan(xy)+z^3\\\mathrm{grad} \ u=\cfrac{\partial u}{\partial x} \ \vec{i}+\cfrac{\partial u}{\partial y}\ \vec{j}+\cfrac{\partial u}{\partial z}\ \vec{k}=\cfrac{y}{1+x^2y^2}\ \vec{i}+\cfrac{x}{1+x^2y^2}\ \vec{j}+3z^2\ \vec{k}\\\mathrm{grad}_B\ u=\cfrac{-1}{1+1}\ \vec{i}+\cfrac{1}{1+1}\ \vec{j}+3\cdot 4\ \vec{k}=-\cfrac{\vec{i}}{2}+\cfrac{\vec{j}}{2}-12\vec{k}\\\cfrac{\partial u}{\partial e}=\nabla u\cdot \cfrac{\vec{a}}{|\vec{a}|}=\mathrm{grad} \ u\cdot \cfrac{\vec{a}}{|\vec{a}|}=$
$=\cfrac{1}{|\vec{a}|}\left(-\cfrac{2y}{1+x^2y^2}\ \vec{i}+\cfrac{3x}{1+x^2y^2}\ \vec{j}+3\sqrt{3}z^2\ \vec{k}\right)\\\left(\cfrac{\partial u}{\partial a}\right)_B=\cfrac{1}{|\vec{a}|}\left(\cfrac{2}{1+1}\ \vec{i}+\cfrac{3}{1+1}\ \vec{j}+3\sqrt{3}\cdot 4\ \vec{k}\right)=\\=\cfrac{1}{|\vec{a}|}\left(\vec{i}+\cfrac{3}{2}\ \vec{j}+12\sqrt{3}\ \vec{k}\right)\\|\vec{a}|=\sqrt{4+9+3}=4\\\left(\cfrac{\partial u}{\partial a}\right)_B=\cfrac{\vec{i}}{4}+\cfrac{3}{8}\ \vec{j}+3\sqrt{3} \ \vec{k}$