例 12. 这样好理解:
记 u = e^xsiny, v = x^2 + y^2, 则 z = f(u, v),
∂z/∂x = (∂f/∂u)(∂u/∂x) + (∂f/∂v)(∂v/∂x) = e^xsiny(∂f/∂u) + 2x(∂f/∂v),
这里 (∂f/∂u)即 f'1, (∂f/∂v)即 f'2.
∂z/∂y = (∂f/∂u)(∂u/∂y) + (∂f/∂v)(∂v/∂y) = e^xcosy(∂f/∂u) + 2y(∂f/∂v).
(∂z/∂x)^2 +(∂z/∂y)^2
= e^(2x)(∂f/∂u)^2 + 2e^x(xsiny+ycosy)(∂f/∂u)(∂f/∂v) + 4(x^2+y^2)(∂f/∂v)^2.
若求二阶偏导数,则
∂^2z/∂x^2 = ∂[e^xsiny(∂f/∂u) + 2x(∂f/∂v)]/∂x , 注意 ∂f/∂u,∂f/∂v 都是 x,y 的二元函数,
= e^xsiny(∂f/∂u) + e^xsiny[(∂^2f/∂u^2)(∂u/∂x)+(∂^2f/∂u∂v)(∂v/∂x)]
+ 2(∂f/∂v) + 2x[(∂^2f/∂v∂u)(∂u/∂x)+(∂^2f/∂v^2)(∂v/∂x)]
= e^xsiny(∂f/∂u) + 2(∂f/∂v) + e^xsiny[e^xsiny(∂^2f/∂u^2)+2x(∂^2f/∂u∂v)]
+ 2x[e^xsiny(∂^2f/∂v∂u)+2y(∂^2f/∂v^2)]
= e^xsiny(∂f/∂u) + 2(∂f/∂v) + (e^xsiny)^2(∂^2f/∂u^2)
+ 4xe^xsiny(∂^2f/∂v∂u) + 4xy(∂^2f/∂v^2)
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