Question

z=ue的u/v次方 u=x^2+y^2 v=xy 求一阶偏导

Answer 1

z=ue^(u/v),其中 u=x^2+y^2,v=xy ,

z' = u'e^(u/v)+ue^(u/v)*(u'v-uv')/v^2

= e^[(x^2+y^2)/(xy)] {2x+(x^2+y^2)[2yx^2-(x^2+y^2)y/(xy)^2]}

= e^[(x^2+y^2)/(xy)] [2x+(x^2+y^2)(x^2-y^2)/(yx^2)]

= e^[(x^2+y^2)/(xy)] [2x+(x^4-y^4)/(yx^2)].

由轮换性,得

z' = e^[(x^2+y^2)/(xy)] [2y+(y^4-x^4)/(xy^2)].