C.10 · Implicit differentiation and the inverse Jacobian
GRE Subject Test · GRE Mathematics Topic 27 3:12 English narration · English + 中文 subtitles burned in
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This unit differentiates a coupled implicit system as a linear system, uses a nonzero Jacobian determinant to justify a local inverse, and distinguishes inverse partial derivatives from scalar reciprocal rules.
本单元把耦合的隐式方程组当作线性方程组求导，用非零的雅可比行列式说明局部逆的存在，并区分逆映射的偏导数与一元的倒数法则。
For a map from u, v to x, y with Jacobian J, small changes satisfy d x, d y equals J times d u, d v.
对从 (u, v) 到 (x, y)、雅可比矩阵为 J 的映射，微小变化满足 (dx, dy) = J(du, dv)。
To find u x with y held fixed, differentiate both equations with respect to x.
要在 y 固定时求 u_x，就对两个方程都关于 x 求导。
The result is a coupled linear system, not two independent scalar rules.
结果是一个耦合的线性方程组，而不是两个独立的一元法则。
If the partials are continuous and the determinant is nonzero, there is a differentiable local inverse, and its derivative is J inverse.
若偏导数连续且行列式非零，就存在可微的局部逆，其导数是 J⁻¹。
Reading off the entries: u x is g v over the determinant, u y is minus f v over it, and so on.
读出各元素：u_x = g_v/det J，u_y = −f_v/det J，依此类推。
Evaluate at the matching point.
在对应的点处求值。
A local inverse need not be global.
局部逆不一定能延拓为全局逆。
For implicit equations, solve the linear system in u x and v x, with the determinant in the unknowns nonzero; a zero determinant leaves the theorem inconclusive.
对隐式方程，解关于 u_x、v_x 的线性方程组，未知量的行列式须非零；行列式为零时定理无法下结论。
The scalar rule 1 over dx du usually fails: for x equals u plus v, y equals u plus 2v, u x is 2, not 1, because v changes to keep y fixed.
一元法则 1/(dx/du) 通常不成立：对 x = u + v，y = u + 2v，u_x = 2 而不是 1，因为要保持 y 不变，v 也随之改变。
At 1, 0 the output is 1, 1, and J has determinant minus 3.
在 (1, 0) 处，输出为 (1, 1)，J 的行列式为 −3。
J stretches a small square into a parallelogram of area 3, the absolute determinant, and reverses its orientation.
J 把一个小正方形拉伸成面积为 3（行列式的绝对值）的平行四边形，并反转其定向。
Inverting gives u x equals 1 third, not 1 half, and v y equals minus 2 thirds.
求逆得 u_x = 1/3（不是 1/2），v_y = −2/3。
Differentiate both equations and state which output is held fixed.
对两个方程都求导，并说明哪个输出保持不变。
A nonzero determinant guarantees a local inverse under the regularity hypotheses; a zero one is not a proof of impossibility.
在正则性条件下，行列式非零保证局部逆存在；行列式为零并不能证明不可能。
Use inverse entries for sensitivities and the absolute determinant for integrals.
灵敏度用逆矩阵的元素，积分用行列式的绝对值。
The Jacobian has determinant minus 3, nonzero, so a local inverse exists.
雅可比行列式为 −3，非零，所以局部逆存在。
Its inverse has rows u x, u y equal to 1 third, 1 third, and v x, v y equal to 1 third, minus 2 thirds, at output 1, 1.
逆矩阵的行在输出 (1, 1) 处为 (u_x, u_y) = (1/3, 1/3) 和 (v_x, v_y) = (1/3, −2/3)。
Multiplying gives the identity.
相乘得单位矩阵。
The determinant is 4 times u squared plus v squared, positive away from the origin, so each such point has a local inverse.
行列式是 4(u² + v²)，在原点以外为正，所以每个这样的点都有局部逆。
But F of u, v equals F of minus u, minus v, so F is not globally injective.
但 F(u, v) = F(−u, −v)，所以 F 不是全局单射。
At the origin the theorem is silent: the cube map is bijective despite a zero derivative.
在原点处定理不下结论：立方映射尽管导数为零，仍是双射。
Check yourself.
自我检测。
For u plus v and u plus 2v, u x is 2.
对 u + v 和 u + 2v，u_x = 2。
A local inverse needs a nonzero determinant; a zero one proves nothing by itself.
局部逆需要行列式非零；行列式为零本身什么也证明不了。
The squaring map is locally but not globally invertible.
平方映射局部可逆，但全局不可逆。
And integrals use the absolute determinant.
积分使用行列式的绝对值。