Question

Задание на фотографии

Answer 1

$dfrac{partial f}{partial x}=-dfrac{ycdot (sqrt{x^2+z^2})'_x}{x^2+z^2}=-dfrac{ycdot 2x}{2sqrt{x^2+z^2}(x^2+z^2)}=-dfrac{xy}{(x^2+z^2)^{3/2}}\ \ \ dfrac{partial f}{partial x}(-1;1;0)=-dfrac{(-1)cdot 1}{((-1)^2+0^2)^{3/2}}=-1$

$dfrac{partial f}{partial y}=dfrac{1}{sqrt{x^2+z^2}}\ \ dfrac{partial f}{partial y}(-1;1;0)=dfrac{1}{sqrt{(-1)^2+0^2}}=1$

$dfrac{partial f}{partial z}=-dfrac{ycdot (sqrt{x^2+z^2})'_z}{x^2+z^2}=-dfrac{ycdot 2z}{2sqrt{x^2+z^2}(x^2+z^2)}=-dfrac{yz}{(x^2+z^2)^{3/2}}\ \ dfrac{partial f}{partial z}(-1;1;0)=-dfrac{1cdot 0}{((-1)^2+0^2)^{3/2}}=0$