Question

Задание на фотографии

Answer 1

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

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

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