Question

13.设z=z(x,y)是由方程 xyz=sin(xyz) 所确定的隐函数,求(OZ)/3oz 1？

Answer 1

根据隐函数求导公式，可以得到：
$\frac{\partial z}{\partial x} = -\frac{yz\cos(xyz)}{xy^2z\cos(xyz)-z^2x\cos(xyz)} = -\frac{y\cos(xyz)}{xy\cos(xyz)-z\cos(xyz)}$

$\frac{\partial z}{\partial y} = -\frac{x\cos(xyz)}{xy^2\cos(xyz)-z^2y\cos(xyz)} = -\frac{x\cos(xyz)}{xy\cos(xyz)-z\cos(xyz)}$

根据全微分的定义，有：

$d(z) = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy$

令$t = xyz$，则有：

$dt = (yzdx + xzdy) + (xydz)$

因此，

$dz = \frac{1}{x^2y^2}\left[(yzdx + xzdy) + (xydz)\right]$

$= \frac{y^2zdx + x^2zdy + xydz}{x^2y^2}$

根据题目所求，有：

$\frac{(OZ)}{3oz} = \frac{\partial z}{\partial x} \cdot \frac{\partial^2 t}{\partial x^2} + 2 \frac{\partial z}{\partial x} \cdot \frac{\partial^2 t}{\partial x \partial y} + \frac{\partial z}{\partial y} \cdot \frac{\partial^2 t}{\partial y^2}$

其中，

$\frac{\partial^2 t}{\partial x^2} = y^2z\sin(xyz) - 2yz^2\cos(xyz) - xz\sin(xyz)$

$\frac{\partial^2 t}{\partial y^2} = x^2z\sin(xyz) - 2xz^2\cos(xyz) - yz\sin(xyz)$

$\frac{\partial^2 t}{\partial x \partial y} = z^2\cos(xyz) - xy\cos(xyz)$

代入上面求得的$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$，并化简可得：

$\frac{(OZ)}{3oz} = -\frac{1}{x^2y^2}\left[(y^3z^2\cos(xyz) - 3xyz^3\cos(xyz) + x^2yz\sin(xyz))\frac{\partial z}{\partial x} + (x^3z^2\cos(xyz) - 3xyz^3\cos(xyz) + y^2xz\sin(xyz))\frac{\partial z}{\partial y}\right]$

代入上面求得的$\frac{\partial z}{\partial x}$和$\frac{\partial z}{\partial y}$的式子，整理可得：

$\frac{(OZ)}{3oz} = -\frac{z\cos(xyz)}{3xy\cos(xyz) - 3z\cos(xyz)}$

因此，$\frac{(OZ)}{3oz}$的值可以根据$x$和$y$的值以及$x,y$对应的$z$值来计算。

Answer 2

xyz = sin(xyz) 两边分别对 x 求偏导数，得
yz + xy∂z/∂x = (yz + xy∂z/∂x) cos(xyz)
∂z/∂x = [yzcos(xyz)-yz]/[xy-xycos(xyz)] = -z/x
由轮换性，得 ∂z/∂y = -z/y