设 z = F ( x + f ( 2 x − y ) , y ) , 其 中 F , f 二 阶 连 续 偏 导 数 ， 求 ∂ 2 z ∂ y 2 . 设z = F(x+f(2x-y),y),其中F,f二阶连续偏导数，求\frac{\partial ^2 z}{\partial y^2}. 设z=F(x+f(2x−y),y),其中F,f二阶连续偏导数，求∂y2∂2z​.

解： ∂ z ∂ y = F 1 ′ f ′ ⋅ ( − 1 ) + F 2 ′ \frac{\partial z}{\partial y} = F'_1f'·(-1)+F'_2 ∂y∂z​=F1′​f′⋅(−1)+F2′​ ∂ 2 z ∂ y 2 = ∂ F 1 ′ f ′ ⋅ ( − 1 ) ∂ y + ∂ F 2 ′ ∂ y \frac{\partial ^2z}{\partial y^2} = \frac{\partial F'_1f'·(-1)}{\partial y} + \frac{\partial F'_2}{\partial y} ∂y2∂2z​=∂y∂F1′​f′⋅(−1)​+∂y∂F2′​​ 其中： ∂ F 1 ′ f ′ ⋅ ( − 1 ) ∂ y = f ′ ( F 11 ′ ′ f ′ − F 12 ′ ′ ) + F 1 ′ f ′ ′ \frac{\partial F'_1f'·(-1)}{\partial y} = f'(F''_{11}f'-F''_{12}) + F'_1f'' ∂y∂F1′​f′⋅(−1)​=f′(F11′′​f′−F12′′​)+F1′​f′′ ∂ F 2 ′ ∂ y = F 21 ′ ′ f ′ ⋅ ( − 1 ) + F 22 ′ ′ \frac{\partial F'_2}{\partial y} = F''_{21}f'·(-1) + F''_{22} ∂y∂F2′​​=F21′′​f′⋅(−1)+F22′′​ 综上： ∂ 2 z ∂ y 2 = F 11 ′ ′ ( f ′ ) 2 − 2 F 12 ′ ′ f ′ + F 1 ′ f ′ ′ + F 22 ′ ′ \frac{\partial ^2z}{\partial y^2} = F''_{11}(f')^2-2F''_{12}f' + F'_1f'' + F''_{22} ∂y2∂2z​=F11′′​(f′)2−2F12′′​f′+F1′​f′′+F22′′​