Inverse function theorem

In mathematics, the inverse function theorem gives sufficient conditions for a vector-valued function to be invertible on an open region containing a point in its ___domain.

The theorem states that if at a point p a function f : Rⁿ → Rⁿ has a Jacobian determinant that is nonzero, and f is continuously differentiable near p, then it is an invertible function near p. That is, an inverse function to f exists in some neighborhood of f(p).

The Jacobian matrix of f⁻¹ at f(p) is then the inverse of Jf, evaluated at p.

The inverse function theorem can be generalized to differentiable maps between differentiable manifolds. In this context the theorem states that for a differentiable map f : M → N, if the derivative of f, (Df)_p : T_pM → T_f(p)N is a linear isomorphism at a point p in M then there exists an open neighborhood U of p such that f|_U : 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.

This can be expressed more clearly as ${\displaystyle f'(a)={{1} \over {(f^{-1})'(f(a))}}}$. Where ' indicates the derivative of the function.

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 ${\displaystyle u=\ln x}$  and restrict the ___domain to ${\displaystyle x>0}$ .

${\displaystyle {\frac {d}{dx}}\ln x={{1} \over {{\frac {d}{du}}e^{u}}}={{1} \over {e^{u}}}={{1} \over {e^{\ln x}}}={{1} \over {x}}}$ 

For more general logarithms, we see that ${\displaystyle {\frac {d}{dx}}\log _{b}(x)={\frac {1}{x\ln(b)}}={\frac {\log _{b}(e)}{x}}}$ .

A similar approach can be used to differentiate an inverse trigonometric function. Let ${\displaystyle u=\tan x}$ .

${\displaystyle {\frac {d}{dx}}\arctan x={{1} \over {{\frac {d}{du}}\tan u}}=\cos ^{2}{u}=\cos ^{2}{\arctan x}=\left({{1} \over {\sqrt {1+x^{2}}}}\right)^{2}={{1} \over {1+x^{2}}}}$