Question

z=tg(xy^2)
Найти:d^2z/dxdy,d^2z/dydx,dz

Answer 1

$z = \mathrm{tg}(xy^2)$

$\frac{\partial z}{\partial x} = \frac{1}{\cos^2(xy^2)}\cdot\frac{\partial (xy^2)}{\partial x} =$

$= \frac{y^2}{\cos^2(xy^2)}$

$\frac{\partial z}{\partial y} = \frac{1}{\cos^2(xy^2)}\cdot\frac{\partial (xy^2)}{\partial y} =$

$= \frac{2xy}{\cos^2(xy^2)}$

$\mathrm{d}z = \frac{\partial z}{\partial x}\cdot\mathrm{d}x + \frac{\partial z}{\partial y}\cdot\mathrm{d}y =$

$= \frac{y^2}{\cos^2(xy^2)}\cdot\mathrm{d}x + \frac{2xy}{\cos^2(xy^2)}\cdot\mathrm{d}y$

$\frac{\partial^2 z}{\partial x\partial y} = \frac{\partial }{\partial x} \frac{2xy}{\cos^2(xy^2)} =$

$= \frac{1}{\cos^4(xy^2)}\cdot\left( 2y\cdot\cos^2(xy^2) - 2xy\cdot 2\cos(xy^2)\cdot(-\sin(xy^2))\cdot y^2 \right) =$

$= \frac{1}{\cos^3(xy^2)}\cdot\left( 2y\cos(xy^2) + 4xy^3\sin(xy^2)\right) =$

$= \frac{ 2y\cdot(\cos(xy^2) + 2xy^2\sin(xy^2) )}{\cos^3(xy^2)}$

$\frac{\partial^2 z}{\partial y\partial x} = \frac{\partial }{\partial y} \frac{y^2}{\cos^2(xy^2)} =$

$= \frac{1}{\cos^4(xy^2)}\cdot\left( 2y\cos^2(xy^2) - y^2\cdot 2\cos(xy^2)\cdot(-\sin(xy^2))\cdot 2xy \right) =$

$= \frac{2y\cdot ( \cos(xy^2) + 2xy^2\sin(xy^2) )}{\cos^3(xy^2)}$