Question

Помогите с решением, высшая математика ​

Answer 1

Объяснение:

$z=\sqrt[3]{x^2+y} \ \ \ \ \ M_0(1;0)\ \ \ \ \ M_1(2,96;-1,12)\\1)\\ \frac{\partial z}{\partial x}= (\sqrt[3]{x^2+y})'_x =((x^2+y)^\frac{1}{3} )'_x=\frac{1}{3 }(x^2+y)^{-\frac{2}{3}}*(x^2+y)'_x=\frac{2x}{3*\sqrt[3]{(x^2+y)^2} } .\\ \frac{\partial z}{\partial y}= (\sqrt[3]{x^2+y})'_y =((x^2+y)^\frac{1}{3} )'_y=\frac{1}{3 }(x^2+y)^{-\frac{2}{3}}*(x^2+y)'_y=\frac{1}{3*\sqrt[3]{(x^2+y)^2} } .\\dz= \frac{2x}{3*\sqrt[3]{(x^2+y)^2} } dx+\frac{1}{3*\sqrt[3]{(x^2+y)^2} } dy.$

$2)\\z=\sqrt[3]{x^2+y} \ \ \ \ M_0(1;0)\\\frac{2*1}{3*\sqrt[3]{1^2+0} }* (x-1)+\frac{1}{(3*\sqrt[3]{1^2+0)} } *(y-0)=\frac{2}{3*1} *(x-1)+\frac{1}{3*1}*y=\\ =\frac{2}{3}*(x-1)+\frac{1}{3}y=\frac{2x-3}{3}+\frac{y}{3}=0 .\\ \frac{2x-2}{3} +\frac{y}{3}=0 \ |*3\\ 2x-2+y=0\\2x+y-2=0.$

$3)\\z=\sqrt[3]{x^2+y}\ \ \ \ (2,96;-1,12)\\ x=2,96 \ \ \ \ \ y=-1,12\\x_0=3\ \ \ \ \ y_0=-1.\\1)\ z(x_0;y_0)=\sqrt[3]{(3^2+(-1)} =\sqrt[3]{9-1} =\sqrt[3]{8}=2.\\ 2)\ \frac{\partial z}{\partial x}\ |_{\left {{x=x_0} \atop {y=y_0} \right.}=\frac{2}{3} .\\ 3)\ \frac{\partial z}{\partial y}\ |_{\left {{x=x_0} \atop {y=y_0} \right.}=\frac{1}{3} .\\z(2,96;-1,12)=2+\frac{2}{3}*(3-2,96)+\frac{1}{3}*(-1,12-(-1))=2+\frac{2*0,04}{3} - \frac{0,12}{3}=$

$=2+\frac{0,08}{3}-\frac{0,12}{3} =2+\frac{0,08-0,12}{3}=2+\frac{-0,112}{3}\approx2-0,037 \approx1,963.$