Question

设f(u,v)为二元可微函数，z=f(x^y,y^x)，求∂z/∂x,∂z/∂y

Answer 1

z = f(x^y, y^x)，记 u = x^y, v = y^x，
则 ∂u/∂x = yx^(y-1), ∂u/∂y = x^ylnx, ∂v/∂x = y^xlny, ∂v/∂y = xy^(x-1).
∂z/∂x = (∂f/∂u)(∂u/∂x) + (∂f/∂v)(∂v/∂x) = yx^(y-1) ∂f/∂u + y^xlny ∂f/∂v
∂z/∂y = (∂f/∂u)(∂u/∂y) + (∂f/∂v)(∂v/∂y) = x^ylnx ∂f/∂u + xy^(x-1) ∂f/∂v

Answer 2

简单分析一下，详情如图所示

Answer 3

∂z/∂x=∂z/∂f[(∂f/∂x)*y*x^(y-1)+∂f/∂y*y^x*lny]
∂z/∂y=∂z/∂f[(∂f/∂x)*x^y*lnx+∂f/∂y*x*y^(x-1)]

Answer 4

限定x,y>0啊，
第一个=f1*y*x^(y-1)+f2*logy*y^x
f1,f2在x^y,y^x取值