设u=e^x.siny,∂u/∂x=e^x.siny,∂u/∂y=e^x.cosy
v=x²+y²，∂v/∂x=2x，∂v/∂y=2y
z=f(u,v)
∂z/∂x=∂f/∂u.∂u/∂x+∂f/∂v.∂v/∂x
=∂f/∂u.e^x.siny+∂f/∂v.2x
=e^x.siny∂f/∂u+2x∂f/∂v
将上面的偏导数对y再求一次偏导数，就得到∂²z/∂x∂y：
∂²z/∂x∂y=e^x.cosy∂f/∂u+e^x.siny[∂²f/∂u².∂u/∂y+∂²f/∂u∂v.∂v/∂y]
+2x[∂²f/∂v∂u.∂u/∂y+∂²f/∂v².∂v/∂y]
=e^x.cosy∂f/∂u+e^x.siny[∂²f/∂u².e^x.cosy+∂²f/∂u∂v.2y]
+2x[∂²f/∂v∂u.e^x.cosy+∂²f/∂v².2y]
=e^x.cosy∂f/∂u+e^2x.sinycosy∂²f/∂u²+2e^x(ysiny+xcosy)∂²f/∂u∂v
+4xy∂²f/∂v²
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