Editing Inverse function rule (section)

==Additional properties==

* [[Integral|Integrating]] this relationship gives

::<math>{f^{-1}}(x)=\int\frac{1}{f'({f^{-1}}(x))}\,{dx} + C.</math>

:This is only useful if the integral exists. In particular we need <math>f'(x)</math> to be non-zero across the range of integration.

:It follows that a function that has a [[continuous function|continuous]] derivative has an inverse in a [[neighbourhood (mathematics)|neighbourhood]] of every point where the derivative is non-zero. This need not be true if the derivative is not continuous.

* Another very interesting and useful property is the following:

::<math> \int f^{-1}(x)\, {dx} = x f^{-1}(x) - F(f^{-1}(x)) + C </math>

:Where <math> F </math> denotes the antiderivative of <math> f </math>.

* The inverse of the derivative of f(x) is also of interest, as it is used in showing the convexity of the [[Legendre transformation|Legendre transform]]. 
Let  <math> z = f'(x)</math>  then we have, assuming <math> f''(x) \neq 0</math>:<math display="block"> \frac{d(f')^{-1}(z)}{dz} = \frac{1}{f''(x)}</math>This can be shown using the previous notation <math> y = f(x)</math>. Then we have:

:<math display="block"> f'(x) = \frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} =  \frac{dy}{dz} f''(x) \Rightarrow \frac{dy}{dz} = \frac{f'(x) }{f''(x)}</math>Therefore:

:<math> \frac{d(f')^{-1}(z)}{dz} = \frac{dx}{dz} = \frac{dy}{dz}\frac{dx}{dy} = \frac{f'(x)}{f''(x)}\frac{1}{f'(x)} = \frac{1}{f''(x)}</math>

By induction, we can generalize this result for any integer <math> n \ge 1</math>, with  <math> z = f^{(n)}(x)</math>, the nth derivative of f(x), and  <math> y = f^{(n-1)}(x)</math>, assuming <math> f^{(i)}(x) \neq 0 \text{ for } 0 < i \le n+1 </math>:

:<math> \frac{d(f^{(n)})^{-1}(z)}{dz} = \frac{1}{f^{(n+1)}(x)}</math>