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C.10 · Implicit differentiation and the inverse Jacobian

GRE · GRE Subject Test · GRE Mathematics · Topic 27

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Scope and prerequisites

Undergraduate GRE preparation. Local objectives within the reviewed ETS scope; this is original teaching, not an official test or score predictor.

Prerequisites: Chain rule, matrix inversion and local inverse hypotheses.

  • Differentiate a coupled implicit system as a linear system
  • Use a nonzero Jacobian determinant to justify a local inverse
  • Distinguish inverse partial derivatives from scalar reciprocal rules

Jacobian matrix 雅可比矩阵: The matrix of first partial derivatives of a vector-valued map.

local inverse 局部逆映射: An inverse defined on neighbourhoods of a particular input and output point.

Vocabulary Train
English
Jacobian matrix/dʒæˈkəʊbɪən ˈmeɪtrɪks/
local inverse/ˈləʊkl ɪnˈvɜːs/
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Choose and justify a method

For a differentiable map F(u,v)=(x,y)=(f(u,v),g(u,v)), its Jacobian is J=[[f_u,f_v],[g_u,g_v]]. Small changes satisfy [dx,dy] (transpose)=J[du,dv] (transpose) to first order. To find u_x while holding y fixed, differentiate both defining equations with respect to x: f_u u_x+f_v v_x=1 and g_u u_x+g_v v_x=0. These are a coupled linear system, not two independent scalar inverse rules.

If f and g are continuously differentiable near the point and det J=f_u g_v−f_v g_u is nonzero there, the inverse-function theorem supplies a differentiable local inverse. Its derivative is J⁻¹=(1/det J)[[g_v,−f_v],[−g_u,f_u]]. Thus u_x=g_v/det J, u_y=−f_v/det J, v_x=−g_u/det J and v_y=f_u/det J. Evaluate every derivative at the corresponding point. A local inverse need not extend to a global one.

For implicit equations H(u,v,x,y)=0 and K(u,v,x,y)=0, differentiate while fixing the requested independent coordinate. Solve [[H_u,H_v],[K_u,K_v]][u_x,v_x] (transpose)=−[H_x,K_x] (transpose). The determinant in the unknown variables u,v must be nonzero to use the usual implicit-function theorem. Signs on the right come from moving known derivatives to the other side. If the determinant vanishes, this theorem is inconclusive; it does not by itself prove no inverse or no implicit solution exists.

The scalar shortcut du/dx=1/(dx/du) holds for a one-variable inverse with nonzero derivative, but usually fails for a coupled system because v changes to keep y fixed. For x=u+v,y=u+2v, J=[[1,1],[1,2]] has determinant 1 and inverse [[2,−1],[−1,1]]. Therefore u_x=2 while 1/f_u=1. An inverse Jacobian transforms differential sensitivities; a change-of-variables integral instead uses the absolute determinant for area or volume scaling, not a selected inverse entry.

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Worked reasoning

Take x=u²+v and y=u−v. At (u,v)=(1,0), the output is (1,1), and J=[[2,1],[1,−1]] has determinant −3. The inverse is [[1/3,1/3],[1/3,−2/3]], so u_x=1/3 and v_y=−2/3 there. Multiplying J by this inverse gives the identity. The reciprocal 1/f_u=1/2 is not u_x because changing u also requires changing v to keep y fixed.

Implicit differentiation and the inverse Jacobian: course example
Original course illustration; its values belong to the worked example, not the later practice.
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Conditions and counterexamples

Differentiate both equations and state which output is held fixed. Nonzero determinant guarantees a local inverse under the regularity hypotheses; zero determinant is not a proof of impossibility. Use an inverse entry for sensitivities and an absolute determinant for integration.

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Guided application

Let $x=u^2+v,y=u-v$. At $(u,v)=(1,0)$, find the inverse Jacobian and the four inverse partial derivatives. Check by multiplication.

Worked solution

The corresponding output is $(1,1)$. The Jacobian is $J=\begin{pmatrix}2&1\\1&-1\end{pmatrix}$ with determinant -3, nonzero. Continuous partials give a differentiable local inverse. Thus

$$J^{-1}=\begin{pmatrix}1/3&1/3\\1/3&-2/3\end{pmatrix}.$$
Read rows as (u_x,u_y) and (v_x,v_y), all at output (1,1). Multiplication gives the identity. In particular $u_x=1/3$, not $1/x_u=1/2$: v changes to keep y fixed.

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Independent transfer

For $F(u,v)=(u^2-v^2,2uv)$, decide whether its derivative is invertible away from the origin and whether F is globally injective on the punctured plane. What does a zero determinant at the origin establish by itself?

Check after attempting

$\det DF=4(u^2+v^2)>0$ away from zero, so the inverse-function theorem supplies a local inverse at each such point. But $F(u,v)=F(-u,-v)$, so it is not globally injective on the punctured plane. A vanishing determinant at zero means this theorem cannot guarantee a differentiable local inverse there. It is not, by itself, a general proof that no local set-theoretic inverse can exist for any map; for example the scalar cube map is bijective despite derivative zero at zero.

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