Question

高等数学偏导数

Answer 1

1. z = ue^v+ve^(-u), u = xy, v = x/y
z'<x> = z'<u>u'<x>e^v + ue^v v'<x> + v'<x>e^(-u) - ve^(-u)u'<x>
= yz'<u>e^v + (1/y)ue^v + (1/y)e^(-u) - yve^(-u)
= y[z'<u>e^v-ve^(-u)] + (1/y)(ue^v+e^(-u)]
z'<y> = z'<u>u'<y>e^v + ue^v v'<y> + v'<y>e^(-u) - ve^(-u)u'<y>
= xz'<u>e^v + (-x/y^2)ue^v + (-x/y^2)e^(-u) - xve^(-u)
= x[z'<u>e^v-ve^(-u)] - (x/y^2)(ue^v+e^(-u)]