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The Jacobian matrix of ''F''<sup>−1</sup> at ''F''(''p'') is then the inverse of the Jacobian of ''F'', evaluated at ''p''. This can be understood as a special case of the [[chain rule]], which states that for [[linear transformations]] ''F'' and ''G'',
:<math>J_{G \circ F} (p) = J_G (F(p)) \cdot J_F (p)</math>
where J denotes the corresponding Jacobian matrix.
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Assume that the inverse function theorem holds at ''F''(''p''). Let <math>G(p) = F^{-1}(p)</math>.
:<math>J_{F^{-1} \circ F} (p) = J_{F^{-1}} (F(p)) \cdot J_F (p)</math>
:<math>J_{I} (p) \cdot (J_F (p))^{-1} = J_{F^{-1}} (F(p)) \cdot J_F (p) \cdot (J_F (p))^{-1}</math>
:<math>I \cdot (J_F (p))^{-1} = J_{F^{-1}} (F(p)) \cdot I</math>
:<math>(J_F (p))^{-1} = J_{F^{-1}} (F(p))</math>
where ''I'' is the [[identity transformation]]. This is often expressed more clearly as the useful single-variable formula,
:<math>f'(x) = {{1} \over {(f^{-1})'(f(x))}}.</math>
The inverse function theorem can be generalized to differentiable maps between [[differentiable manifold]]s. In this context the theorem states that for a differentiable map ''F'' : ''M'' → ''N'', if the [[pushforward|derivative]] of ''F'',
:(''DF'')<sub>''p''</sub> : T<sub>''p''</sub>''M'' → T<sub>''F''(''p'')</sub>''N'' is a linear isomorphism at a point ''p'' in ''M'' then there exists an open neighborhood ''U'' of ''p'' such that :''F''|<sub>''U''</sub> : ''U'' → ''F''(''U'') is a [[diffeomorphism]]. Note that this implies that ''M'' and ''N'' must have the same dimension. If the derivative of ''F'' is an isomorphism at all points ''p'' in ''M'' then the map ''F'' is a [[local diffeomorphism]].
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===Examples===
Several functions exist for which differentiating the inverse is much easier than differentiating the function itself. Using the inverse function theorem, a derivative of a function's inverse indicates the derivative of the original function. Perhaps the most well-known example is the method used to compute the derivative of the [[natural logarithm]], whose inverse is the [[exponential function]]. Let <math>u = \ln x</math> and restrict the ___domain to <math>x > 0</math>. Then
:<math>\frac{d}{dx}\ln x = {{1} \over {\frac{d}{du}e^u}} = {{1} \over {e^u}} = {{1} \over {e^{\ln x}}} = {{1} \over {x}}.</math>
For more general [[logarithms]], we see that <math>\frac{d}{dx} \log_b(x) = \frac{1}{x \ln(b)} = \frac{\log_b(e)}{x}.</math>
A similar approach can be used to differentiate an inverse [[trigonometric function]]. Let <math>u = \tan x.</math>
:<math>\frac{d}{dx}\arctan x = {{1} \over {\frac{d}{du}\tan u}} = \cos^2{u} = \cos^2{\arctan x} = \left({{1} \over {\sqrt{1+x^2}}}\right)^2 = {{1} \over {1+x^2}}.</math>
[[Category:Multivariate calculus]]
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