Question

1.设 z=sin(xy^2+1) ,求dz

Answer 1

别忘了采纳建议

要求 dz，需要对 z = sin(xy^2 + 1) 求偏导数，然后乘以对应自变量的偏导数。

令 u = xy^2 + 1，则 z = sin(u)，根据链式法则，有：dz/dx = dz/du × du/dx

dz/dy = dz/du × du/dy

对于第一个式子，有：dz/du = cos(u)   du/dx = y^2

因此，dz/dx = dz/du × du/dx = cos(u) × y^2 = y^2 cos(xy^2 + 1)

对于第二个式子，有：dz/du = cos(u)    du/dy = 2xy

因此，dz/dy = dz/du × du/dy = cos(u) × 2xy = 2xy cos(xy^2 + 1)

综上所述，dz/dx = y^2 cos(xy^2 + 1)，dz/dy = 2xy cos(xy^2 + 1)。

Answer 2

z=sin(xy^2+1)
dz
= dsin(xy^2+1)
=cos(xy^2+1) d(xy^2+1)
=cos(xy^2+1) (y^2 dx + 2xy dy)