设Z=f(x^2 +y,2xy),求dz/dx和dz/dy
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设Z=f(x^2+y,2xy),求dz/dx和dz/dy设Z=f(x^2+y,2xy),求dz/dx和dz/dy设Z=f(x^2+y,2xy),求dz/dx和dz/dyu=x^2+y∂u/
设Z=f(x^2 +y,2xy),求dz/dx和dz/dy
设Z=f(x^2 +y,2xy),求dz/dx和dz/dy
设Z=f(x^2 +y,2xy),求dz/dx和dz/dy
u = x^2 + y
∂u/∂x = 2x ∂u/∂y = 1
du = (∂u/∂x)dx + (∂u/∂y)dy = 2xdx + dy
v = 2xy
∂v/∂x = 2y ∂v/∂y = 2x
dv = (∂v/∂x)dx + (∂v/∂y)dy = 2ydx + 2xdy
z = f(u,v)
dz = (∂f/∂u)du + (∂f/∂v)dv
= (∂f/∂u)(2xdx + dy) + (∂f/∂v)(2ydx + 2xdy)
= (2x(∂f/∂u) + 2y(∂f/∂v))dx + ((∂f/∂u)+2x(∂f/∂v))dy
因此你要求的应该是
∂z/∂x = 2x(∂f/∂u) + 2y(∂f/∂v)
∂z/∂y = (∂f/∂u)+2x(∂f/∂v)
根据复合函数的链式求导法则可得
dz/dx=df/dx•2x+df/dy•2y
dz/dy=df/dx+df/dy•2x
由于输入麻烦，此处d均表示求偏导
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