Z=tg (xy^2) найти dz/dy d^2z/dx^2 d^2z/d^2y

Решите задачу:

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

$\dfrac{\partial z}{\partial y}=\dfrac{1}{\cos^2(xy^2)} \cdot(xy^2)'_y=\dfrac{1}{\cos^2(xy^2)} \cdot2xy=\dfrac{2xy}{\cos^2(xy^2)}$

$\dfrac{\partial^2z}{\partial y^2}=\dfrac{(2xy)'_y\cdot\cos^2(xy^2)-2xy\cdot(\cos^2(xy^2))'_y}{(\cos^2(xy^2))^2} =\\\\=\dfrac{2x\cdot\cos^2(xy^2)-2xy\cdot 2\cos(xy^2)\cdot(\cos(xy^2))'_y}{\cos^4(xy^2)} =\\\\=\dfrac{2x\cos^2(xy^2)-4xy\cos(xy^2)\cdot(-\sin(xy^2))\cdot(xy^2)'_y}{\cos^4(xy^2)} =\\\\=\dfrac{2x\cos^2(xy^2)+4xy\cos(xy^2)\sin(xy^2)\cdot2xy}{\cos^4(xy^2)} =\dfrac{2x\cos(xy^2)+8x^2y^2\sin(xy^2)}{\cos^3(xy^2)}$

$\dfrac{\partial z}{\partial x}=\dfrac{1}{\cos^2(xy^2)} \cdot(xy^2)'_x=\dfrac{1}{\cos^2(xy^2)} \cdot y^2=\dfrac{y^2}{\cos^2(xy^2)}$

$\dfrac{\partial^2z}{\partial x^2}=\dfrac{(y^2)'_x\cdot\cos^2(xy^2)-y^2\cdot(\cos^2(xy^2))'_x}{(\cos^2(xy^2))^2} =\\\\=\dfrac{0\cdot\cos^2(xy^2)-y^2\cdot2\cos(xy^2)\cdot(\cos(xy^2))'_x}{\cos^4(xy^2)} =\\\\=\dfrac{-2y^2\cos(xy^2)\cdot(-\sin(xy^2))\cdot(xy^2)'_x}{\cos^4(xy^2)} =\\\\=\dfrac{2y^2\cos(xy^2)\sin(xy^2)\cdot y^2}{\cos^4(xy^2)} =\dfrac{2y^4\sin(xy^2)}{\cos^3(xy^2)}$