Question

u = sqrt(z) * sin(y/x) ;du| M 0 (2:0:4) =

Answer 1

Ответ:

Найти дифференциал функции в точке М₀  .

Формула :   $\bf du=u'_{x}\, dx+u'_{y}\, dy+u'_{z}\, dz\ \ ,\ \ u=u(x,y,z)$  .

$\bf u=\sqrt{z}\cdot sin\dfrac{y}{x}\ \ ,\ \ \ M_0(2;0;4)\\\\u'_{x}=\sqrt{z}\cdot cos\dfrac{y}{x}\cdot \dfrac{-1}{x^2}\ \ ,\ \ u'_{x}(M_0)=-2\cdot cos0\cdot \dfrac{1}{4}=-\dfrac{1}{2}\\\\u'_{y}=\sqrt{z}\cdot cos\dfrac{y}{x}\cdot \dfrac{1}{x}\ \ ,\ \ \ \ u'_{y}(M_0)=2\cdot cos0\cdot \dfrac{1}{2}=1\\\\u'_{z}=\dfrac{1}{2\sqrt{z}}\cdot sin\dfrac{y}{x}\ \ ,\ \ \ \ \ \ \ u'_{z}(M_0)=\dfrac{1}{4}\cdot sin0=0$  

$\bf du=-\sqrt{z}\cdot cos\dfrac{y}{x}\cdot \dfrac{1}{x^2}\cdot dx+\sqrt{z}\cdot cos\dfrac{y}{x}\cdot \dfrac{1}{x}\cdot dy+\dfrac{1}{2\sqrt{z}}\cdot sin\dfrac{y}{x}\cdot dz\\\\\\du\, \Big|_{M_0}=-\dfrac{1}{2}\, dx+dy$