Question

найти градиент grad u(M) функции u = x^(-2)+3xy^2 в точке M(1;0)

Answer 1

Решение во вложении.

Answer 2

$grad\, u(x,y)=\frac{\partial u}{\partial x}\cdot \vec {i}+\frac{\partial u}{\partial y}\cdot \vec {j}\\\\u=x^{-2}+3xy^2\; ,\; \; M(1;0)\\\\\\\frac{\partial u}{\partial x}=-2\cdot x^{-1}+3y^2=-\frac{2}{x}+3y^2\; ,\; \; \frac{\partial u}{\partial x}\, \Big |_{M(1,0)}=-1+0=-2\\\\\frac{\partial u}{\partial y}=0+3x\cdot 2y=6xy\; ,\; \; \; \; \; \frac{\partial u}{\partial y}\, \Big |_{M(1,0)}=6\cdot 1\cdot 0=0\\\\\\grad\, u(1,0)=-2\cdot \vec{i}+0\cdot \vec{j}=-2\cdot \vec{i}$