Question

Найти производную функции u=ln(3-x^2)+xy^2z в точке M1 (1,3,2), по направлению от этой точки к точке M2(0,5,0)

Answer 1

$u=ln(3-x^2)+xy^2z\ \ ,\ \ \ M_1(1;3;2)\ ,\ M_2(0;5;0)\\\\u'_{x}=\dfrac{-2x}{3-x^2}+y^2z\ \ ,\ \ \ u'_{y}=2xyz\ \ ,\ \ \ u'_{z}=xy^2\\\\u'(M_1)=\dfrac{-2}{3-1}+9\cdot 2=17\\\\u'_{y}(M_1)=2\cdot 3\cdot 2=12\\\\u'_{z}=1\cdot 9=9\\\\\vec{l}=\overline{M_1M_2}=(-1;2;-2)\ \ ,\ \ |\overline{M_1M_2}|=\sqrt{1+4+4}=\sqrt9=3\ \ ,$

$\overline{M_1M_2}^\circ =\Big(-\dfrac{1}{3}\, ;\,\dfrac{2}{3}\, ;-\dfrac{2}{3}\Big)\ \ \Rightarrow \ \ cos\alpha =-\dfrac{1}{3} \ ,\ \ cos\beta =\dfrac{2}{3}\ \ ,\ \ cos\gamma =-\dfrac{2}{3}$  

$\dfrac{\partial u}{\partial \vec{l}}\Big|_{M_1}=u'_{x}\Big|_{M_1}\cdot cos\alpha +u'_{y}\Big|_{M_1}\cdot cos\beta +u'_{z}\Big|_{M_1}\cdot cos\gamma \\\\\\\dfrac{\partial u}{\partial \vec{l}}\Big|_{M_1}=-17\cdot \dfrac{1}{3}+12\cdot \dfrac{2}{3}-9\cdot \dfrac{2}{3}=\dfrac{-17+24-18}{3}=-\dfrac{11}{3}=-3\dfrac{2}{3}$