Question

Найти производную функции u в точке М0 по по направлению вектора

Answer 1

Ответ:

$u=x^2y^2z-ln(z-1)\ \ ,\ \ M_0(1;12)\ \ ,\ \ \vec{l}=5\vec{i}-6\vec{j}+2\sqrt5\vec{k}\ \ ,\\\\\\u'_{x}=2xy^2z\ \ ,\ \ \ u'_{y}=2x^2yz\ \ ,\ \ \ u'_{z}=x^2+y^2-\dfrac{1}{z-1}\\\\\\u'_{x}\Big|_{M_0}=2\cdot 2=4\ \ ,\ \ u'_{y}\Big|_{M_0}=2\cdot 2=4\ \ ,\ \ u'_{z}\Big|_{M_0}=1+1-1=1\\\\\\|\, \vec{l}\, |=\sqrt{5^2+6^2+4\cdot 5}=\sqrt{81}=9\\\\cos\alpha =\dfrac{5}{9}\ \ ,\ \ cos\beta =-\dfrac{2}{3}\ \ ,\ \ cos\gamma =\dfrac{2\sqrt5}{9}$

$\dfrac{\partial u}{\partial \vec{l}}\Big|_{M_0}=4\cdot \dfrac{5}{9}-4\cdot \dfrac{2}{3}+1\cdot \dfrac{2\sqrt5}{9}=\dfrac{20-24+2\sqrt5}{9}=\dfrac{2\sqrt5-4}{9}$