求函数u=x+sin(y/2)+eyz的全微分为____．
du=dx+[(1/2)cos(y/2)+zeyz)dy+yeyzdz。解析：因为∂u/∂x=1，∂u/∂y=(1/2)cos(y/2)+zeyz，∂u/∂z=yeyz，所以du=dx+[(1/2)cos(y/2)+zeyz]dy+yeyzdz.
求函数u=x+sin(y/2)+eyz的全微分为____．
du=dx+[(1/2)cos(y/2)+zeyz)dy+yeyzdz。解析：因为∂u/∂x=1，∂u/∂y=(1/2)cos(y/2)+zeyz，∂u/∂z=yeyz，所以du=dx+[(1/2)cos(y/2)+zeyz]dy+yeyzdz.
