# Inverse function rule

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vte :
In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function f in terms of the derivative of f. More precisely, if the inverse of $f$ is denoted as $f^{-1}$, where $f^{-1}(y)=x$ if and only if $f(x)=y$, then the inverse function rule is, in Lagrange's notation,
    $\left[f^{-1}\right]'(a)={\frac {1}{f'\left(f^{-1}(a)\right)}}$.
This formula holds in general whenever $f$ is continuous and injective on an interval I, with $f$ being differentiable at $f^{-1}(a)$($\in I$) and where$f'(f^{-1}(a))\neq 0$. The same formula is also equivalent to the expression
    ${\mathcal {D}}\left[f^{-1}\right]={\frac {1}{({\mathcal {D}}f)\circ \left(f^{-1}\right)}},$
where ${\mathcal {D}}$ denotes the unary derivative operator (on the space of functions) and $\circ$ denotes function composition.
Geometrically, a function and inverse function have graphs that are reflections, in the line $y=x$. This reflection operation turns the gradient of any line into its reciprocal.
Assuming that $f$ has an inverse in a neighbourhood of $x$ and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at $x$ and have a derivative given by the above formula.
The inverse function rule may also be expressed in Leibniz's notation. As that notation suggests,
    ${\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}=1.$
This relation is obtained by differentiating the equation $f^{-1}(y)=x$ in terms of x and applying the chain rule, yielding that:
    ${\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}={\frac {dx}{dx}}$
considering that the derivative of x with respect to x is 1.

## Derivation

Let $f$ be an invertible (bijective) function, let $x$ be in the domain of $f$, and let $y$ be in the codomain of $f$. Since f is a bijective function, $y$ is in the range of $f$. This also means that $y$ is in the domain of $f^{-1}$, and that $x$ is in the codomain of $f^{-1}$. Since $f$ is an invertible function, we know that $f(f^{-1}(y))=y$. The inverse function rule can be obtained by taking the derivative of this equation.
    ${\dfrac {\mathrm {d} }{\mathrm {d} y}}f(f^{-1}(y))={\dfrac {\mathrm {d} }{\mathrm {d} y}}y$
The right side is equal to 1 and the chain rule can be applied to the left side:
    ${\begin{aligned}{\dfrac {\mathrm {d} \left(f(f^{-1}(y))\right)}{\mathrm {d} \left(f^{-1}(y)\right)}}{\dfrac {\mathrm {d} \left(f^{-1}(y)\right)}{\mathrm {d} y}}&=1\\{\dfrac {\mathrm {d} f(f^{-1}(y))}{\mathrm {d} f^{-1}(y)}}{\dfrac {\mathrm {d} f^{-1}(y)}{\mathrm {d} y}}&=1\\f^{\prime }(f^{-1}(y))(f^{-1})^{\prime }(y)&=1\end{aligned}}$
Rearranging then gives
    $(f^{-1})^{\prime }(y)={\frac {1}{f^{\prime }(f^{-1}(y))}}$
Rather than using $y$ as the variable, we can rewrite this equation using $a$ as the input for $f^{-1}$, and we get the following:
    $(f^{-1})^{\prime }(a)={\frac {1}{f^{\prime }\left(f^{-1}(a)\right)}}$

## Examples

* $y=x^{2}$ (for positive x) has inverse $x={\sqrt {y}}$.
    ${\frac {dy}{dx}}=2x{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{2{\sqrt {y}}}}={\frac {1}{2x}}$
    ${\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=2x\cdot {\frac {1}{2x}}=1.$
At $x=0$, however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.
* $y=e^{x}$ (for real x) has inverse $x=\ln {y}$ (for positive $y$)
    ${\frac {dy}{dx}}=e^{x}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{y}}=e^{-x}$
    ${\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=e^{x}\cdot e^{-x}=1.$

## Additional properties

* Integrating this relationship gives
       ${f^{-1}}(x)=\int {\frac {1}{f'({f^{-1}}(x))}}\,{dx}+C.$
    This is only useful if the integral exists. In particular we need $f'(x)$ to be non-zero across the range of integration.
    It follows that a function that has a continuous derivative has an inverse in a 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:
       $\int f^{-1}(x)\,{dx}=xf^{-1}(x)-F(f^{-1}(x))+C$
    Where $F$ denotes the antiderivative of $f$.
* The inverse of the derivative of f(x) is also of interest, as it is used in showing the convexity of the Legendre transform.
Let $z=f'(x)$ then we have, assuming $f''(x)\neq 0$:${\frac {d(f')^{-1}(z)}{dz}}={\frac {1}{f''(x)}}$This can be shown using the previous notation $y=f(x)$. Then we have:
    $f'(x)={\frac {dy}{dx}}={\frac {dy}{dz}}{\frac {dz}{dx}}={\frac {dy}{dz}}f''(x)\Rightarrow {\frac {dy}{dz}}={\frac {f'(x)}{f''(x)}}$Therefore:
    ${\frac {d(f')^{-1}(z)}{dz}}={\frac {dx}{dz}}={\frac {dy}{dz}}{\frac {dx}{dy}}={\frac {f'(x)}{f''(x)}}{\frac {1}{f'(x)}}={\frac {1}{f''(x)}}$
By induction, we can generalize this result for any integer $n\geq 1$, with $z=f^{(n)}(x)$, the nth derivative of f(x), and $y=f^{(n-1)}(x)$, assuming $f^{(i)}(x)\neq 0{\text{ for }}0<i\leq n+1$:
    ${\frac {d(f^{(n)})^{-1}(z)}{dz}}={\frac {1}{f^{(n+1)}(x)}}$

## Higher derivatives

The chain rule given above is obtained by differentiating the identity $f^{-1}(f(x))=x$ with respect to x. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to x, one obtains
    ${\frac {d^{2}y}{dx^{2}}}\,\cdot \,{\frac {dx}{dy}}+{\frac {d}{dx}}\left({\frac {dx}{dy}}\right)\,\cdot \,\left({\frac {dy}{dx}}\right)=0,$
that is simplified further by the chain rule as
    ${\frac {d^{2}y}{dx^{2}}}\,\cdot \,{\frac {dx}{dy}}+{\frac {d^{2}x}{dy^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{2}=0.$
Replacing the first derivative, using the identity obtained earlier, we get
    ${\frac {d^{2}y}{dx^{2}}}=-{\frac {d^{2}x}{dy^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{3}.$
Similarly for the third derivative:
    ${\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{4}-3{\frac {d^{2}x}{dy^{2}}}\,\cdot \,{\frac {d^{2}y}{dx^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{2}$
or using the formula for the second derivative,
    ${\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{4}+3\left({\frac {d^{2}x}{dy^{2}}}\right)^{2}\,\cdot \,\left({\frac {dy}{dx}}\right)^{5}$
These formulas are generalized by the Faà di Bruno's formula.
These formulas can also be written using Lagrange's notation. If f and g are inverses, then
    $g''(x)={\frac {-f''(g(x))}{[f'(g(x))]^{3}}}$

## Example

* $y=e^{x}$ has the inverse $x=\ln y$. Using the formula for the second derivative of the inverse function,
    ${\frac {dy}{dx}}={\frac {d^{2}y}{dx^{2}}}=e^{x}=y{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}\left({\frac {dy}{dx}}\right)^{3}=y^{3};$
so that
    ${\frac {d^{2}x}{dy^{2}}}\,\cdot \,y^{3}+y=0{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {d^{2}x}{dy^{2}}}=-{\frac {1}{y^{2}}}$,
which agrees with the direct calculation.