Question

43 баллов даю. Только 9 нужно решить, остальные не надо. ​

Answer 1

$z=tg(xy^2)\\\\z'_{x}=\frac{1}{cos^2(xy^2)}\cdot y^2\; \; ,\; \; \; z'_{y}=\frac{1}{cos^2(xy^2)}\cdot 2xy\\\\\\z''_{xx}=\frac{-y^2\, \cdot \, 2cos(xy^2)\cdot (-sin(xy^2))\cdot y^2}{cos^4(xy^2)}=\frac{y^4\, \cdot \, sin(2xy^2)}{cos^4(xy^2)}\\\\\\z''_{yy}=\frac{2x\cdot cos^2(xy^2)-2xy\, \cdot \, 2cos(xy^2)\cdot (-sin(xy^2))\cdot 2xy}{cos^4(xy^2)}=\frac{2x\cdot cos^2(xy^2)-4x^2y^2\cdot sin(2xy^2)}{cos^4(xy^2)}$

$z''_{xy}=\frac{2y\cdot cos^2(xy^2)-y^2\cdot 2cos(xy^2)\cdot (-sin(xy^2))\cdot 2xy}{cos^4(xy^2)}=\frac{2y\cdot cos^2(xy^2)+2xy^3\cdot sin(2xy^2)}{cos^4(xy^2)}\\\\\\z''_{yx}=\frac{2y\cdot cos^2(xy^2)-2xy\cdot 2cos(xy^2)\cdot (-sin(xy^2))\cdot y^2}{cos^4(xy^2)}=\frac{2y\cdot cos^2(xy^2)+2xy^3\cdot sin(2xy^2)}{cos^4(xy^2)}\\\\z''_{xy}=z''_{yx}$