Question

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

Answer 1

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

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

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

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

$dfrac{partial^2z}{partial x^2}=dfrac{(y^2)'_xcdotcos^2(xy^2)-y^2cdot(cos^2(xy^2))'_x}{(cos^2(xy^2))^2} =\\=dfrac{0cdotcos^2(xy^2)-y^2cdot2cos(xy^2)cdot(cos(xy^2))'_x}{cos^4(xy^2)} =\\=dfrac{-2y^2cos(xy^2)cdot(-sin(xy^2))cdot(xy^2)'_x}{cos^4(xy^2)} =\\=dfrac{2y^2cos(xy^2)sin(xy^2)cdot y^2}{cos^4(xy^2)} =dfrac{2y^4sin(xy^2)}{cos^3(xy^2)}$