高等数学（第七版）同济大学 习题9-4 （后2题）个人解答
高等数学（第七版）同济大学 习题9-4（后2题）
12. 求下列函数的 ∂ 2 z ∂ x 2 ， ∂ 2 z ∂ x ∂ y ， ∂ 2 z ∂ y 2 （其中 f 具有二阶连续偏导数）： \begin{aligned}&12. \ 求下列函数的\frac{\partial^2 z}{\partial x^2}，\frac{\partial^2 z}{\partial x \partial y}，\frac{\partial^2 z}{\partial y^2}（其中f具有二阶连续偏导数）：&\end{aligned} 12. 求下列函数的∂x2∂2z，∂x∂y∂2z，∂y2∂2z（其中f具有二阶连续偏导数）：
( 1 ) z = f ( x y , y ) ； ( 2 ) z = f ( x , x y ) ； ( 3 ) z = f ( x y 2 , x 2 y ) ； ( 4 ) z = f ( s i n x , c o s y , e x + y ) . \begin{aligned} &\ \ (1)\ \ z=f(xy, \ y)；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)\ \ z=f\left(x, \ \frac{x}{y}\right)；\\\\ &\ \ (3)\ \ z=f(xy^2, \ x^2y)；\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)\ \ z=f(sin\ x, \ cos\ y, \ e^{x+y}). & \end{aligned} (1) z=f(xy, y)； (2) z=f(x, yx)； (3) z=f(xy2, x2y)； (4) z=f(sin x, cos y, ex+y).
解：
( 1 ) 令 s = x y ， t = y ，则 z = f ( s , t ) ， s 和 t 为中间变量，将 s ， t 编号 1 ， 2 ，则 ∂ z ∂ x = f 1 ′ ⋅ ∂ s ∂ x = y f 1 ′ ， ∂ z ∂ y = f 1 ′ ⋅ ∂ s ∂ y + f 2 ′ ⋅ d t d y = x f 1 ′ + f 2 ′ ，因为 f ( s , t ) 是 s 和 t 的函数，所以 f 1 ′ 和 f 2 ′ 也是 s 和 t 的函数， f 1 ′ 和 f 2 ′ 是以 s 和 t 为中间变量的 x 和 y 的函数，所以 ∂ 2 z ∂ x 2 = ∂ ∂ x ( ∂ z ∂ x ) = ∂ ∂ x ( y f 1 ′ ) = y f 11 ′ ′ ⋅ ∂ s ∂ x = y 2 f 11 ′ ′ ， ∂ 2 z ∂ x ∂ y = ∂ ∂ y ( ∂ z ∂ x ) = ∂ ∂ y ( y f 1 ′ ) = f 1 ′ + y ( f 11 ′ ′ ⋅ ∂ s ∂ y + f 12 ′ ′ ⋅ d t d y ) = f 1 ′ + x y f 11 ′ ′ + y f 12 ′ ′ ， ∂ 2 z ∂ y 2 = ∂ ∂ y ( ∂ z ∂ y ) = ∂ ∂ y ( x f 1 ′ + f 2 ′ ) = x ( f 11 ′ ′ ∂ s ∂ y + f 12 ′ ′ d t d y ) + f 21 ′ ′ ∂ s ∂ y + f 22 ′ ′ d t d y = x 2 f 11 ′ ′ + 2 x f 12 ′ ′ + f 22 ′ ′ . ( 2 ) 令 s = x ， t = x y ，将 s 和 t 编号 1 ， 2 ，则 ∂ z ∂ x = f 1 ′ d s d x + f 2 ′ ∂ t ∂ x = f 1 ′ + 1 y f 2 ′ ， ∂ z ∂ y = f 2 ′ ∂ t ∂ y = − x y 2 f 2 ′ ， 因为 f ( s , t ) 是 s 和 t 的函数，所以 f 1 ′ 和 f 2 ′ 也是 s 和 t 的函数， f 1 ′ 和 f 2 ′ 是以 s 和 t 为中间变量的 x 和 y 的函数，所以 ∂ 2 z ∂ x 2 = ∂ ∂ x ( ∂ z ∂ x ) = ∂ ∂ x ( f 1 ′ + 1 y f 2 ′ ) = f 11 ′ ′ + f 12 ′ ′ ⋅ ∂ t ∂ x + 1 y ( f 21 ′ ′ + f 22 ′ ′ ⋅ ∂ t ∂ x ) = f 11 ′ ′ + 2 y f 12 ′ ′ + 1 y 2 f 22 ′ ′ ， ∂ 2 z ∂ x ∂ y = ∂ ∂ y ( ∂ z ∂ x ) = ∂ ∂ y ( f 1 ′ + 1 y f 2 ′ ) = f 12 ′ ′ ⋅ ∂ t ∂ y − 1 y 2 f 2 ′ + 1 y f 22 ′ ′ ∂ t ∂ y = − x y 2 f 12 ′ ′ − 1 y 2 f 2 ′ − x y 3 f 22 ′ ′ ， ∂ 2 z ∂ y 2 = ∂ ∂ y ( ∂ z ∂ y ) = ∂ ∂ y ( − x y 2 f 2 ′ ) = 2 x y 3 f 2 ′ − x y 2 f 22 ′ ′ ∂ t ∂ v = 2 x y 3 f 2 ′ + x 2 y 4 f 22 ′ ′ . ( 3 ) 令 s = x y 2 ， t = x 2 y ，将 s ， t 编号为 1 ， 2 ，则 ∂ z ∂ x = f 1 ′ ∂ s ∂ x + f 2 ′ ∂ t ∂ x = y 2 f 1 ′ + 2 x y f 2 ′ ， ∂ z ∂ y = f 1 ′ ∂ s ∂ y + f 2 ′ ∂ t ∂ y = 2 x y f 1 ′ + x 2 f 2 ′ ， ∂ 2 z ∂ x 2 = ∂ ∂ x ( ∂ z ∂ x ) = ∂ ∂ x ( y 2 f 1 ′ + 2 x y f 2 ′ ) = y 2 ( f 11 ′ ′ ⋅ ∂ s ∂ x + f 12 ′ ′ ⋅ ∂ t ∂ x ) + 2 y f 2 ′ + 2 x y ( f 21 ′ ′ ∂ s ∂ x + f 22 ′ ′ ∂ t ∂ x ) = y 2 ( y 2 f 11 ′ ′ + 2 x y f 12 ′ ′ ) + 2 y f 2 ′ + 2 x y ( y 2 f 21 ′ ′ + 2 x y f 22 ′ ′ ) = 2 y f 2 ′ + y 4 f 11 ′ ′ + 4 x y 3 f 12 ′ ′ + 4 x 2 y 2 f 22 ′ ′ ， ∂ 2 z ∂ x ∂ y = ∂ ∂ y ( ∂ z ∂ x ) = ∂ ∂ y ( y 2 f 1 ′ + 2 x y f 2 ′ ) = 2 y f 1 ′ + y 2 ( f 11 ′ ′ ⋅ ∂ s ∂ y + f 12 ′ ′ ⋅ ∂ t ∂ y ) + 2 x f 2 ′ + 2 x y ( f 21 ′ ′ ∂ s ∂ y + f 22 ′ ′ ∂ t ∂ y ) = 2 y f 1 ′ + y 2 ( 2 x y f 11 ′ ′ + x 2 f 12 ′ ′ ) + 2 x f 2 ′ + 2 x y ( 2 x y f 21 ′ ′ + x 2 f 22 ′ ′ ) = 2 y f 1 ′ + 2 x f 2 ′ + 2 x y 3 f 11 ′ ′ + 5 x 2 y 2 f 12 ′ ′ + 2 x 3 y f 22 ′ ′ ， ∂ 2 z ∂ y 2 = ∂ ∂ y ( ∂ z ∂ y ) = ∂ ∂ y ( 2 x y f 1 ′ + x 2 f 2 ′ ) = 2 x f 1 ′ + 2 x y ( f 11 ′ ′ ∂ s ∂ y + f 12 ′ ′ ∂ t ∂ y ) + x 2 ( f 21 ′ ′ ∂ s ∂ y + f 22 ′ ′ ∂ t ∂ y ) = 2 x f 1 ′ + 2 x y ( 2 x y f 11 ′ ′ + x 2 f 12 ′ ′ ) + x 2 ( 2 x y f 21 ′ ′ + x 2 f 22 ′ ′ ) = 2 x f 1 ′ + 4 x 2 y 2 f 11 ′ ′ + 4 x 3 y f 12 ′ ′ + x 4 f 22 ′ ′ . ( 4 ) 令 u = s i n x ， v = c o s y ， w = e x + y ，将 u ， v ， w 编号为 1 ， 2 ， 3 ，则 ∂ z ∂ x = f 1 ′ d u d x + f 3 ′ ∂ w ∂ x = c o s x f 1 ′ + e x + y f 3 ′ ， ∂ z ∂ y = f 2 ′ d v d y + f 3 ′ ∂ w ∂ y = − s i n y f 2 ′ + e x + y f 3 ′ ， ∂ 2 z ∂ x 2 = ∂ ∂ x ( ∂ z ∂ x ) = ∂ ∂ x ( c o s x f 1 ′ + e x + y f 3 ′ ) = − s i n x f 1 ′ + c o s x ( f 11 ′ ′ d u d x + f 13 ′ ′ ∂ w ∂ x ) + e x + y f 3 ′ + e x + y ( f 31 ′ ′ d u d x + f 33 ′ ′ ∂ w ∂ x ) = − s i n x f 1 ′ + c o s x ( c o s x f 11 ′ ′ + e x + y f 13 ′ ′ ) + e x + y f 3 ′ + e x + y ( c o s x f 31 ′ ′ + e x + y f 33 ′ ′ ) = e x + y f 3 ′ − s i n x f 1 ′ + c o s 2 x f 11 ′ ′ + 2 e x + y c o s x f 13 ′ ′ + e 2 ( x + y ) f 33 ′ ′ ， ∂ 2 z ∂ x ∂ y = ∂ ∂ y ( ∂ z ∂ x ) = ∂ ∂ y ( c o s x f 1 ′ + e x + y f 3 ′ ) = c o s x ( f 12 ′ ′ d v d y + f 13 ′ ′ ∂ w ∂ y ) + e x + y f 3 ′ + e x + y ( f 32 ′ ′ d v d y + f 33 ′ ′ ∂ w ∂ y ) = c o s x ( − s i n y f 12 ′ ′ + e x + y f 13 ′ ′ ) + e x + y f 3 ′ + e x + y ( − s i n y f 32 ′ ′ + e x + y f 33 ′ ′ ) = e x + y f 3 ′ − c o s x s i n y f 12 ′ ′ + e x + y c o s x f 13 ′ ′ − e x + y s i n y f 32 ′ ′ + e 2 ( x + y ) f 33 ′ ′ ， ∂ 2 z ∂ y 2 = ∂ ∂ y ( ∂ z ∂ y ) = ∂ ∂ y ( − s i n y f 2 ′ + e x + y f 3 ′ ) = − c o s y f 2 ′ − s i n y ( f 22 ′ ′ d v d y + f 23 ′ ′ ∂ w ∂ y ) + e x + y f 3 ′ + e x + y ( f 32 ′ ′ d v d y + f 33 ′ ′ ∂ w ∂ y ) = − c o s y f 2 ′ − s i n y ( − s i n y f 22 ′ ′ + e x + y f 23 ′ ′ ) + e x + y f 3 ′ + e x + y ( − s i n y f 32 ′ ′ + e x + y f 33 ′ ′ ) = e x + y f 3 ′ − c o s y f 2 ′ + s i n 2 y f 22 ′ ′ − 2 e x + y s i n y f 23 ′ ′ + e 2 ( x + y ) f 33 ′ ′ . \begin{aligned} &\ \ (1)\ 令s=xy，t=y，则z=f(s, \ t)，s和t为中间变量，将s，t编号1，2，则\frac{\partial z}{\partial x}=f'_1\cdot \frac{\partial s}{\partial x}=yf'_1，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial z}{\partial y}=f'_1\cdot \frac{\partial s}{\partial y}+f'_2\cdot \frac{dt}{dy}=xf'_1+f'_2，因为f(s, \ t)是s和t的函数，所以f'_1和f'_2也是s和t的函数，f'_1和f'_2是以s和t\\\\ &\ \ \ \ \ \ \ \ 为中间变量的x和y的函数，所以\frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial x}(yf'_1)=yf''_{11}\cdot \frac{\partial s}{\partial x}=y^2f''_{11}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial y}(yf'_1)=f'_1+y\left(f''_{11}\cdot \frac{\partial s}{\partial y}+f''_{12}\cdot \frac{dt}{dy}\right)=f'_1+xyf''_{11}+yf''_{12}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial}{\partial y}(xf'_1+f'_2)=x\left(f''_{11}\frac{\partial s}{\partial y}+f''_{12}\frac{dt}{dy}\right)+f''_{21}\frac{\partial s}{\partial y}+f''_{22}\frac{dt}{dy}=x^2f''_{11}+2xf''_{12}+f''_{22}.\\\\ &\ \ (2)\ 令s=x，t=\frac{x}{y}，将s和t编号1，2，则\frac{\partial z}{\partial x}=f'_1\frac{ds}{dx}+f'_2\frac{\partial t}{\partial x}=f'_1+\frac{1}{y}f'_2，\frac{\partial z}{\partial y}=f'_2\frac{\partial t}{\partial y}=-\frac{x}{y^2}f'_2，\\\\ &\ \ \ \ \ \ \ \ 因为f(s, \ t)是s和t的函数，所以f'_1和f'_2也是s和t 的函数，f'_1和f'_2是以s和t为中间变量的x和y的函数，所以\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial x}\left(f'_1+\frac{1}{y}f'_2\right)=f''_{11}+f''_{12}\cdot \frac{\partial t}{\partial x}+\frac{1}{y}\left(f''_{21}+f''_{22}\cdot \frac{\partial t}{\partial x}\right)=f''_{11}+\frac{2}{y}f''_{12}+\frac{1}{y^2}f''_{22}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial y}\left(f'_1+\frac{1}{y}f'_2\right)=f''_{12}\cdot \frac{\partial t}{\partial y}-\frac{1}{y^2}f'_2+\frac{1}{y}f''_{22}\frac{\partial t}{\partial y}=-\frac{x}{y^2}f''_{12}-\frac{1}{y^2}f'_2-\frac{x}{y^3}f''_{22}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial}{\partial y}\left(-\frac{x}{y^2}f'_2\right)=\frac{2x}{y^3}f'_2-\frac{x}{y^2}f''_{22}\frac{\partial t}{\partial v}=\frac{2x}{y^3}f'_2+\frac{x^2}{y^4}f''_{22}.\\\\ &\ \ (3)\ 令s=xy^2，t=x^2y，将s，t编号为1，2，则\frac{\partial z}{\partial x}=f'_1\frac{\partial s}{\partial x}+f'_2\frac{\partial t}{\partial x}=y^2f'_1+2xyf'_2，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial z}{\partial y}=f'_1\frac{\partial s}{\partial y}+f'_2\frac{\partial t}{\partial y}=2xyf'_1+x^2f'_2，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial x}(y^2f'_1+2xyf'_2)=y^2\left(f''_{11}\cdot \frac{\partial s}{\partial x}+f''_{12}\cdot \frac{\partial t}{\partial x}\right)+2yf'_2+2xy\left(f''_{21}\frac{\partial s}{\partial x}+f''_{22}\frac{\partial t}{\partial x}\right)=\\\\ &\ \ \ \ \ \ \ \ y^2(y^2f''_{11}+2xyf''_{12})+2yf'_2+2xy(y^2f''_{21}+2xyf''_{22})=2yf'_2+y^4f''_{11}+4xy^3f''_{12}+4x^2y^2f''_{22}，\\\\ &\ \ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial y}(y^2f'_1+2xyf'_2)=2yf'_1+y^2\left(f''_{11}\cdot \frac{\partial s}{\partial y}+f''_{12}\cdot \frac{\partial t}{\partial y}\right)+2xf'_2+2xy\left(f''_{21}\frac{\partial s}{\partial y}+f''_{22}\frac{\partial t}{\partial y}\right)=\\\\ &\ \ \ \ \ \ \ \ 2yf'_1+y^2(2xyf''_{11}+x^2f''_{12})+2xf'_2+2xy(2xyf''_{21}+x^2f''_{22})=2yf'_1+2xf'_2+2xy^3f''_{11}+5x^2y^2f''_{12}+2x^3yf''_{22}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial}{\partial y}(2xyf'_1+x^2f'_2)=2xf'_1+2xy\left(f''_{11}\frac{\partial s}{\partial y}+f''_{12}\frac{\partial t}{\partial y}\right)+x^2\left(f''_{21}\frac{\partial s}{\partial y}+f''_{22}\frac{\partial t}{\partial y}\right)=\\\\ &\ \ \ \ \ \ \ \ 2xf'_1+2xy(2xyf''_{11}+x^2f''_{12})+x^2(2xyf''_{21}+x^2f''_{22})=2xf'_1+4x^2y^2f''_{11}+4x^3yf''_{12}+x^4f''_{22}.\\\\ &\ \ (4)\ 令u=sin\ x，v=cos\ y，w=e^{x+y}，将u，v，w编号为1，2，3，则\\\\ &\ \ \ \ \ \ \ \ \frac{\partial z}{\partial x}=f'_1\frac{du}{dx}+f'_3\frac{\partial w}{\partial x}=cos\ xf'_1+e^{x+y}f'_3，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial z}{\partial y}=f'_2\frac{dv}{dy}+f'_3\frac{\partial w}{\partial y}=-sin\ yf'_2+e^{x+y}f'_3，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial x}(cos\ xf'_1+e^{x+y}f'_3)=\\\\ &\ \ \ \ \ \ \ \ -sin\ xf'_1+cos\ x\left(f''_{11}\frac{du}{dx}+f''_{13}\frac{\partial w}{\partial x}\right)+e^{x+y}f'_3+e^{x+y}\left(f''_{31}\frac{du}{dx}+f''_{33}\frac{\partial w}{\partial x}\right)=\\\\ &\ \ \ \ \ \ \ \ -sin\ xf'_1+cos\ x(cos\ xf''_{11}+e^{x+y}f''_{13})+e^{x+y}f'_3+e^{x+y}(cos\ xf''_{31}+e^{x+y}f''_{33})=\\\\ &\ \ \ \ \ \ \ \ e^{x+y}f'_3-sin\ xf'_1+cos^2 xf''_{11}+2e^{x+y}cos\ xf''_{13}+e^{2(x+y)}f''_{33}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial x \partial y}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)=\frac{\partial}{\partial y}(cos\ xf'_1+e^{x+y}f'_3)=cos\ x\left(f''_{12}\frac{dv}{dy}+f''_{13}\frac{\partial w}{\partial y}\right)+e^{x+y}f'_3+e^{x+y}\left(f''_{32}\frac{dv}{dy}+f''_{33}\frac{\partial w}{\partial y}\right)=\\\\ &\ \ \ \ \ \ \ \ cos\ x(-sin\ yf''_{12}+e^{x+y}f''_{13})+e^{x+y}f'_3+e^{x+y}(-sin\ yf''_{32}+e^{x+y}f''_{33})=\\\\ &\ \ \ \ \ \ \ \ e^{x+y}f'_3-cos\ xsin\ yf''_{12}+e^{x+y}cos\ xf''_{13}-e^{x+y}sin\ yf''_{32}+e^{2(x+y)}f''_{33}，\\\\ &\ \ \ \ \ \ \ \ \frac{\partial^2 z}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)=\frac{\partial}{\partial y}(-sin\ yf'_2+e^{x+y}f'_3)=\\\\ &\ \ \ \ \ \ \ \ -cos\ yf'_2-sin\ y\left(f''_{22}\frac{dv}{dy}+f''_{23}\frac{\partial w}{\partial y}\right)+e^{x+y}f'_3+e^{x+y}\left(f''_{32}\frac{dv}{dy}+f''_{33}\frac{\partial w}{\partial y}\right)=\\\\ &\ \ \ \ \ \ \ \ -cos\ yf'_2-sin\ y(-sin\ yf''_{22}+e^{x+y}f''_{23})+e^{x+y}f'_3+e^{x+y}(-sin\ yf''_{32}+e^{x+y}f''_{33})=\\\\ &\ \ \ \ \ \ \ \ e^{x+y}f'_3-cos\ yf'_2+sin^2 yf''_{22}-2e^{x+y}sin\ yf''_{23}+e^{2(x+y)}f''_{33}. & \end{aligned} (1) 令s=xy，t=y，则z=f(s, t)，s和t为中间变量，将s，t编号1，2，则∂x∂z=f1′⋅∂x∂s=yf1′， ∂y∂z=f1′⋅∂y∂s+f2′⋅dydt=xf1′+f2′，因为f(s, t)是s和t的函数，所以f1′和f2′也是s和t的函数，f1′和f2′是以s和t 为中间变量的x和y的函数，所以∂x2∂2z=∂x∂(∂x∂z)=∂x∂(yf1′)=yf11′′⋅∂x∂s=y2f11′′， ∂x∂y∂2z=∂y∂(∂x∂z)=∂y∂(yf1′)=f1′+y(f11′′⋅∂y∂s+f12′′⋅dydt)=f1′+xyf11′′+yf12′′， ∂y2∂2z=∂y∂(∂y∂z)=∂y∂(xf1′+f2′)=x(f11′′∂y∂s+f12′′dydt)+f21′′∂y∂s+f22′′dydt=x2f11′′+2xf12′′+f22′′. (2) 令s=x，t=yx，将s和t编号1，2，则∂x∂z=f1′dxds+f2′∂x∂t=f1′+y1f2′，∂y∂z=f2′∂y∂t=−y2xf2′， 因为f(s, t)是s和t的函数，所以f1′和f2′也是s和t的函数，f1′和f2′是以s和t为中间变量的x和y的函数，所以 ∂x2∂2z=∂x∂(∂x∂z)=∂x∂(f1′+y1f2′)=f11′′+f12′′⋅∂x∂t+y1(f21′′+f22′′⋅∂x∂t)=f11′′+y2f12′′+y21f22′′， ∂x∂y∂2z=∂y∂(∂x∂z)=∂y∂(f1′+y1f2′)=f12′′⋅∂y∂t−y21f2′+y1f22′′∂y∂t=−y2xf12′′−y21f2′−y3xf22′′， ∂y2∂2z=∂y∂(∂y∂z)=∂y∂(−y2xf2′)=y32xf2′−y2xf22′′∂v∂t=y32xf2′+y4x2f22′′. (3) 令s=xy2，t=x2y，将s，t编号为1，2，则∂x∂z=f1′∂x∂s+f2′∂x∂t=y2f1′+2xyf2′， ∂y∂z=f1′∂y∂s+f2′∂y∂t=2xyf1′+x2f2′， ∂x2∂2z=∂x∂(∂x∂z)=∂x∂(y2f1′+2xyf2′)=y2(f11′′⋅∂x∂s+f12′′⋅∂x∂t)+2yf2′+2xy(f21′′∂x∂s+f22′′∂x∂t)= y2(y2f11′′+2xyf12′′)+2yf2′+2xy(y2f21′′+2xyf22′′)=2yf2′+y4f11′′+4xy3f12′′+4x2y2f22′′， ∂x∂y∂2z=∂y∂(∂x∂z)=∂y∂(y2f1′+2xyf2′)=2yf1′+y2(f11′′⋅∂y∂s+f12′′⋅∂y∂t)+2xf2′+2xy(f21′′∂y∂s+f22′′∂y∂t)= 2yf1′+y2(2xyf11′′+x2f12′′)+2xf2′+2xy(2xyf21′′+x2f22′′)=2yf1′+2xf2′+2xy3f11′′+5x2y2f12′′+2x3yf22′′， ∂y2∂2z=∂y∂(∂y∂z)=∂y∂(2xyf1′+x2f2′)=2xf1′+2xy(f11′′∂y∂s+f12′′∂y∂t)+x2(f21′′∂y∂s+f22′′∂y∂t)= 2xf1′+2xy(2xyf11′′+x2f12′′)+x2(2xyf21′′+x2f22′′)=2xf1′+4x2y2f11′′+4x3yf12′′+x4f22′′. (4) 令u=sin x，v=cos y，w=ex+y，将u，v，w编号为1，2，3，则 ∂x∂z=f1′dxdu+f3′∂x∂w=cos xf1′+ex+yf3′， ∂y∂z=f2′dydv+f3′∂y∂w=−sin yf2′+ex+yf3′， ∂x2∂2z=∂x∂(∂x∂z)=∂x∂(cos xf1′+ex+yf3′)= −sin xf1′+cos x(f11′′dxdu+f13′′∂x∂w)+ex+yf3′+ex+y(f31′′dxdu+f33′′∂x∂w)= −sin xf1′+cos x(cos xf11′′+ex+yf13′′)+ex+yf3′+ex+y(cos xf31′′+ex+yf33′′)= ex+yf3′−sin xf1′+cos2xf11′′+2ex+ycos xf13′′+e2(x+y)f33′′， ∂x∂y∂2z=∂y∂(∂x∂z)=∂y∂(cos xf1′+ex+yf3′)=cos x(f12′′dydv+f13′′∂y∂w)+ex+yf3′+ex+y(f32′′dydv+f33′′∂y∂w)= cos x(−sin yf12′′+ex+yf13′′)+ex+yf3′+ex+y(−sin yf32′′+ex+yf33′′)= ex+yf3′−cos xsin yf12′′+ex+ycos xf13′′−ex+ysin yf32′′+e2(x+y)f33′′， ∂y2∂2z=∂y∂(∂y∂z)=∂y∂(−sin yf2′+ex+yf3′)= −cos yf2′−sin y(f22′′dydv+f23′′∂y∂w)+ex+yf3′+ex+y(f32′′dydv+f33′′∂y∂w)= −cos yf2′−sin y(−sin yf22′′+ex+yf23′′)+ex+yf3′+ex+y(−sin yf32′′+ex+yf33′′)= ex+yf3′−cos yf2′+sin2yf22′′−2ex+ysin yf23′′+e2(x+y)f33′′.
13. 设 u = f ( x , y ) 的所有二阶偏导数连续，而 x = s − 3 t 2 ， y = 3 s + t 2 ，证明 ( ∂ u ∂ x ) 2 + ( ∂ u ∂ y ) 2 = ( ∂ u ∂ s ) 2 + ( ∂ u ∂ t ) 2 及 ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = ∂ 2 u ∂ s 2 + ∂ 2 u ∂ t 2 . \begin{aligned}&13. \ 设u=f(x, \ y)的所有二阶偏导数连续，而x=\frac{s-\sqrt{3}t}{2}，y=\frac{\sqrt{3}s+t}{2}，证明\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2=\\\\&\ \ \ \ \ \ \left(\frac{\partial u}{\partial s}\right)^2+\left(\frac{\partial u}{\partial t}\right)^2及\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial s^2}+\frac{\partial^2 u}{\partial t^2}.&\end{aligned} 13. 设u=f(x, y)的所有二阶偏导数连续，而x=2s−3t，y=23s+t，证明(∂x∂u)2+(∂y∂u)2= (∂s∂u)2+(∂t∂u)2及∂x2∂2u+∂y2∂2u=∂s2∂2u+∂t2∂2u.
解：
∂ u ∂ s = ∂ u ∂ x ∂ x ∂ s + ∂ u ∂ y ∂ y ∂ s = 1 2 ∂ u ∂ x + 3 2 ∂ u ∂ y ， ∂ u ∂ t = ∂ u ∂ x ∂ x ∂ t + ∂ u ∂ y ∂ y ∂ t = − 3 2 ∂ u ∂ x + 1 2 ∂ u ∂ y ， 因此， ( ∂ u ∂ s ) 2 + ( ∂ u ∂ t ) 2 = ( 1 2 ∂ u ∂ x + 3 2 ∂ u ∂ y ) 2 + ( − 3 2 ∂ u ∂ x + 1 2 ∂ u ∂ y ) 2 = ( ∂ u ∂ x ) 2 + ( ∂ u ∂ y ) 2 又因 ∂ 2 u ∂ s 2 = ∂ ∂ s ( ∂ u ∂ s ) = ∂ ∂ s ( 1 2 ∂ u ∂ x + 3 2 ∂ u ∂ y ) = 1 2 ( ∂ 2 u ∂ x 2 ∂ x ∂ s + ∂ 2 u ∂ x ∂ y ∂ y ∂ s ) + 3 2 ( ∂ 2 u ∂ y ∂ x ∂ x ∂ s + ∂ 2 u ∂ y 2 ∂ y ∂ s ) = 1 2 ( 1 2 ∂ 2 u ∂ x 2 + 3 2 ∂ 2 u ∂ x ∂ y ) + 3 2 ( 1 2 ∂ 2 u ∂ y ∂ x + 3 2 ∂ 2 u ∂ y 2 ) = 1 4 ∂ 2 u ∂ x 2 + 3 2 ∂ 2 u ∂ x ∂ y + 3 4 ∂ 2 u ∂ y 2 ， ∂ 2 u ∂ t 2 = ∂ ∂ t ( ∂ u ∂ t ) = ∂ ∂ t ( − 3 2 ∂ u ∂ x + 1 2 ∂ u ∂ y ) = − 3 2 ( ∂ 2 u ∂ x 2 ∂ x ∂ t + ∂ 2 u ∂ x ∂ y ∂ y ∂ t ) + 1 2 ( ∂ 2 u ∂ y ∂ x ∂ x ∂ t + ∂ 2 u ∂ y 2 ∂ y ∂ t ) = − 3 2 ( − 3 2 ∂ 2 u ∂ x 2 + 1 2 ∂ 2 u ∂ x ∂ y ) + 1 2 ( − 3 2 ∂ 2 u ∂ y ∂ x + 1 2 ∂ 2 u ∂ y 2 ) = 3 4 ∂ 2 u ∂ x 2 − 3 2 ∂ 2 u ∂ x ∂ y + 1 4 ∂ 2 u ∂ y 2 ， 所以， ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = ∂ 2 u ∂ s 2 + ∂ 2 u ∂ t 2 \begin{aligned} &\ \ \frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}=\frac{1}{2}\frac{\partial u}{\partial x}+\frac{\sqrt{3}}{2}\frac{\partial u}{\partial y}，\\\\ &\ \ \frac{\partial u}{\partial t}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial t}=-\frac{\sqrt{3}}{2}\frac{\partial u}{\partial x}+\frac{1}{2}\frac{\partial u}{\partial y}，\\\\ &\ \ 因此，\left(\frac{\partial u}{\partial s}\right)^2+\left(\frac{\partial u}{\partial t}\right)^2=\left(\frac{1}{2}\frac{\partial u}{\partial x}+\frac{\sqrt{3}}{2}\frac{\partial u}{\partial y}\right)^2+\left(-\frac{\sqrt{3}}{2}\frac{\partial u}{\partial x}+\frac{1}{2}\frac{\partial u}{\partial y}\right)^2=\left(\frac{\partial u}{\partial x}\right)^2+\left(\frac{\partial u}{\partial y}\right)^2\\\\ &\ \ 又因\frac{\partial^2 u}{\partial s^2}=\frac{\partial}{\partial s}\left(\frac{\partial u}{\partial s}\right)=\frac{\partial}{\partial s}\left(\frac{1}{2}\frac{\partial u}{\partial x}+\frac{\sqrt{3}}{2}\frac{\partial u}{\partial y}\right)=\frac{1}{2}\left(\frac{\partial^2 u}{\partial x^2}\frac{\partial x}{\partial s}+\frac{\partial^2 u}{\partial x\partial y}\frac{\partial y}{\partial s}\right)+\frac{\sqrt{3}}{2}\left(\frac{\partial^2 u}{\partial y\partial x}\frac{\partial x}{\partial s}+\frac{\partial^2 u}{\partial y^2}\frac{\partial y}{\partial s}\right)=\\\\ &\ \ \frac{1}{2}\left(\frac{1}{2}\frac{\partial^2 u}{\partial x^2}+\frac{\sqrt{3}}{2}\frac{\partial^2 u}{\partial x\partial y}\right)+\frac{\sqrt{3}}{2}\left(\frac{1}{2}\frac{\partial^2 u}{\partial y\partial x}+\frac{\sqrt{3}}{2}\frac{\partial^2 u}{\partial y^2}\right)=\frac{1}{4}\frac{\partial^2 u}{\partial x^2}+\frac{\sqrt{3}}{2}\frac{\partial^2 u}{\partial x\partial y}+\frac{3}{4}\frac{\partial^2 u}{\partial y^2}，\\\\ &\ \ \frac{\partial^2 u}{\partial t^2}=\frac{\partial}{\partial t}\left(\frac{\partial u}{\partial t}\right)=\frac{\partial}{\partial t}\left(-\frac{\sqrt{3}}{2}\frac{\partial u}{\partial x}+\frac{1}{2}\frac{\partial u}{\partial y}\right)=-\frac{\sqrt{3}}{2}\left(\frac{\partial^2 u}{\partial x^2}\frac{\partial x}{\partial t}+\frac{\partial^2 u}{\partial x\partial y}\frac{\partial y}{\partial t}\right)+\frac{1}{2}\left(\frac{\partial^2 u}{\partial y\partial x}\frac{\partial x}{\partial t}+\frac{\partial^2 u}{\partial y^2}\frac{\partial y}{\partial t}\right)=\\\\ &\ \ -\frac{\sqrt{3}}{2}\left(-\frac{\sqrt{3}}{2}\frac{\partial^2 u}{\partial x^2}+\frac{1}{2}\frac{\partial^2 u}{\partial x\partial y}\right)+\frac{1}{2}\left(-\frac{\sqrt{3}}{2}\frac{\partial^2 u}{\partial y\partial x}+\frac{1}{2}\frac{\partial^2 u}{\partial y^2}\right)=\frac{3}{4}\frac{\partial^2 u}{\partial x^2}-\frac{\sqrt{3}}{2}\frac{\partial^2 u}{\partial x\partial y}+\frac{1}{4}\frac{\partial^2 u}{\partial y^2}，\\\\ &\ \ 所以，\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial s^2}+\frac{\partial^2 u}{\partial t^2} & \end{aligned} ∂s∂u=∂x∂u∂s∂x+∂y∂u∂s∂y=21∂x∂u+23∂y∂u， ∂t∂u=∂x∂u∂t∂x+∂y∂u∂t∂y=−23∂x∂u+21∂y∂u， 因此，(∂s∂u)2+(∂t∂u)2=(21∂x∂u+23∂y∂u)2+(−23∂x∂u+21∂y∂u)2=(∂x∂u)2+(∂y∂u)2 又因∂s2∂2u=∂s∂(∂s∂u)=∂s∂(21∂x∂u+23∂y∂u)=21(∂x2∂2u∂s∂x+∂x∂y∂2u∂s∂y)+23(∂y∂x∂2u∂s∂x+∂y2∂2u∂s∂y)= 21(21∂x2∂2u+23∂x∂y∂2u)+23(21∂y∂x∂2u+23∂y2∂2u)=41∂x2∂2u+23∂x∂y∂2u+43∂y2∂2u， ∂t2∂2u=∂t∂(∂t∂u)=∂t∂(−23∂x∂u+21∂y∂u)=−23(∂x2∂2u∂t∂x+∂x∂y∂2u∂t∂y)+21(∂y∂x∂2u∂t∂x+∂y2∂2u∂t∂y)= −23(−23∂x2∂2u+21∂x∂y∂2u)+21(−23∂y∂x∂2u+21∂y2∂2u)=43∂x2∂2u−23∂x∂y∂2u+41∂y2∂2u， 所以，∂x2∂2u+∂y2∂2u=∂s2∂2u+∂t2∂2u