Inverse function rule

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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 ${\displaystyle f}$ is denoted as ${\displaystyle f^{-1$ , where ${\displaystyle f^{-1}(y)=x}$ if and only if ${\displaystyle f(x)=y}$ , then the inverse function rule is, in Lagrange's notation,

${\displaystyle \left[f^{-1}\right]'(y)={\frac {1}{f'\left(f^{-1}(y)\right).}$

This formula holds in general whenever ${\displaystyle f}$ is continuous and injective on an interval I, with ${\displaystyle f}$ being differentiable at ${\displaystyle f^{-1}(y)}$ ( ${\displaystyle \in I}$ ) and where ${\displaystyle f'(f^{-1}(y))\neq 0}$ . The same formula is also equivalent to the expression

${\displaystyle {\mathcal {D\left[f^{-1}\right]={\frac {1}{({\mathcal {Df)\circ \left(f^{-1}\right),}$

where ${\displaystyle {\mathcal {D$ denotes the unary derivative operator (on the space of functions) and ${\displaystyle \circ }$ denotes function composition.

Geometrically, a function and inverse function have graphs that are reflections, in the line ${\displaystyle y=x}$ . This reflection operation turns the gradient of any line into its reciprocal.^[1]

Assuming that ${\displaystyle f}$ has an inverse in a neighbourhood of ${\displaystyle x}$ and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at ${\displaystyle x}$ 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,

${\displaystyle {\frac {dx}{dy\,{\frac {dy}{dx=1.}$

This relation is obtained by differentiating the equation ${\displaystyle f^{-1}(y)=x}$ in terms of x and applying the chain rule, yielding that:

${\displaystyle {\frac {dx}{dy\,{\frac {dy}{dx={\frac {dx}{dx$

considering that the derivative of x with respect to x is 1.

Derivation#edit

Let ${\displaystyle f}$ be an invertible (bijective) function, let ${\displaystyle x}$ be in the domain of ${\displaystyle f}$ , and let ${\displaystyle y=f(x).}$ Let ${\displaystyle g=f^{-1}.}$ So, ${\displaystyle f(g(y))=y.}$ Differentiating this equation with respect to ⁠ ${\displaystyle y}$ ⁠, and using the chain rule, one gets

${\displaystyle f'(g(y))\cdot g'(y)=1.}$

That is,

${\displaystyle g'(y)={\frac {1}{f'(g(y))$

${\displaystyle \left[f^{-1}\right]^{\prime }(y)={\frac {1}{f^{\prime }(f^{-1}(y)).}$

Examples#edit

${\displaystyle y=x^{2$ (for positive x) has inverse ${\displaystyle x={\sqrt {y$ .

${\displaystyle {\frac {dy}{dx=2x{\mbox{ {\mbox{ {\mbox{ {\mbox{ ;{\mbox{ {\mbox{ {\mbox{ {\mbox{ {\frac {dx}{dy={\frac {1}{2{\sqrt {y&#61$

${\displaystyle {\frac {dy}{dx\,{\frac {dx}{dy=2x\cdot {\frac {1}{2x=1.}$

At ${\displaystyle x=0}$ , however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.

${\displaystyle y=e^{x$ (for real x) has inverse ${\displaystyle x=\ln {y$ (for positive ${\displaystyle y}$ )

${\displaystyle {\frac {dy}{dx=e^{x}{\mbox{ {\mbox{ {\mbox{ {\mbox{ ;{\mbox{ {\mbox{ {\mbox{ {\mbox{ {\frac {dx}{dy={\frac {1}{y=e^&#1$

${\displaystyle {\frac {dy}{dx\,{\frac {dx}{dy=e^{x}e^{-x}=1.}$

Additional properties#edit

Integrating this relationship gives

${\displaystyle {f^{-1(y)=\int {\frac {1}{f'({f^{-1(y))\,{dy}+C.}$

This is only useful if the integral exists. In particular we need ${\displaystyle f'(x)}$ 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:

${\displaystyle \int f^{-1}(y)\,{dy}=yf^{-1}(y)-F(f^{-1}(y))+C}$

where ${\displaystyle F}$ denotes the antiderivative of ${\displaystyle f}$ .

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 ${\displaystyle z=f'(x)}$ then we have, assuming ${\displaystyle f''(x)\neq 0}$ : ${\displaystyle {\frac {d}{dz\left[f'\right]^{-1}(z)={\frac {1}{f''(x)$ This can be shown using the previous notation ${\displaystyle y=f(x)}$ . Then we have:

${\displaystyle f'(x)={\frac {dy}{dx={\frac {dy}{dz{\frac {dz}{dx={\frac {dy}{dzf''(x)\Rightarrow {\frac {dy}{dz={\frac {f'(x)}{f''(x)$ Therefore:

${\displaystyle {\frac {d}{dz[f']^{-1}(z)={\frac {dx}{dz={\frac {dy}{dz{\frac {dx}{dy={\frac {f'(x)}{f''(x){\frac {1}{f'(x)={\frac &#$

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

${\displaystyle {\frac {d}{dz\left[f^{(n)}\right]^{-1}(z)={\frac {1}{f^{(n+1)}(x)$

Higher order derivatives#edit

The chain rule given above is obtained by differentiating the identity ${\displaystyle f^{-1}(y)=x}$ with respect to y, where ${\displaystyle y=f(x)}$ . One can continue the same process for higher derivatives. Differentiating the identity twice with respect to x, one obtains

${\displaystyle {\frac {d^{2}y}{dx^{2\,{\frac {dx}{dy+{\frac {d}{dx\left({\frac {dx}{dy\right)\,\left({\frac {dy}{dx\right)=0,}$

that is simplified further by the chain rule as

${\displaystyle {\frac {d^{2}y}{dx^{2\,{\frac {dx}{dy+{\frac {d^{2}x}{dy^{2\,\left({\frac {dy}{dx\right)^{2}=0.}$

Replacing the first derivative, using the identity obtained earlier, we get

${\displaystyle {\frac {d^{2}y}{dx^{2=-{\frac {d^{2}x}{dy^{2\,\left({\frac {dy}{dx\right)^{3$

which implies

${\displaystyle {\frac {d^{2}x}{dy^{2=-{\frac {d^{2}y/dx^{2{\left(dy/dx\right)^{3.}$

Similarly for the third derivative we have

${\displaystyle {\frac {d^{3}y}{dx^{3=-{\frac {d^{3}x}{dy^{3\,\left({\frac {dy}{dx\right)^{4}-3{\frac {d^{2}x}{dy^{2\,{\frac {d^{2}y&#125$

Using the formula for the second derivative, we get

${\displaystyle {\frac {d^{3}y}{dx^{3=-{\frac {d^{3}x}{dy^{3\,\left({\frac {dy}{dx\right)^{4}+3\left({\frac {d^{2}y}{dx^{2\right)^{2}\,\left({$

which implies

${\displaystyle {\frac {d^{3}x}{dy^{3=-{\frac {d^{3}y/dx^{3{\left(dy/dx\right)^{4+3{\frac {\left(d^{2}y/dx^{2}\right)^{2{\left(dy/dx\right)^{5.}$

In general, higher order derivatives of an inverse function can be expressed with Faà di Bruno's formula. Alternatively, the nth derivative can be written succinctly as an application of a differential operator:

${\displaystyle \left[f^{-1}\right]^{(n)}(y)=\left.\left[\left({\frac {1}{f'(t){\frac {d}{dt\right)^{n}t\right]\right|_{t=f^{-1}(y)}.}$

In Lagrange's notation, the second and third derivatives from the examples above can be found as applying the operator and substituting ${\displaystyle t=f^{-1}(y)}$ after:

${\displaystyle \left[f^{-1}\right]''(y)=\left.\left[{\frac {1}{f'(t){\frac {d}{dt\left({\frac {1}{f'(t){\frac {d}{dtt\right)\right]\right|_{t=f^{-1}(y)}$

${\displaystyle \left[f^{-1}\right]'''(y)=\left.\left[{\frac {1}{f'(t){\frac {d}{dt\left({\frac {1}{f'(t){\frac {d}{dt\left({\frac {1}{f'(t){\frac {d}&$

One can also derive the nth-integration of inverse function with base-point a using Cauchy formula for repeated integration whenever ${\displaystyle f(f^{-1}(y))=y}$ :

${\displaystyle \left[f^{-1}\right]^{(-n)}(y)={\frac {1}{n!\left(f^{-1}(a)(y-a)^{n}+\int _{f^{-1}(a)}^{f^{-1}(y)}\left(y-f(u)\right)^{n}\,du\right).}$

Example#edit

${\displaystyle y=e^{x$ has the inverse ${\displaystyle x=\ln y}$ . Using the formula for the second derivative of the inverse function,

${\displaystyle {\frac {dy}{dx={\frac {d^{2}y}{dx^{2=e^{x}=y{\mbox{ {\mbox{ {\mbox{ {\mbox{ ;{\mbox{ {\mbox{ {\mbox{ {\mbox{ \left({\f$

so that

${\displaystyle {\frac {d^{2}x}{dy^{2\,\cdot \,y^{3}+y=0{\mbox{ {\mbox{ {\mbox{ {\mbox{ ;{\mbox{ {\mbox{ {\mbox{ {\mbox{ {\frac {d^{2}x}{$

which agrees with the direct calculation.

References#edit

Citations#edit

^ "Derivatives of Inverse Functions". oregonstate.edu. Archived from the original on 2021-04-10. Retrieved 2019-07-26.

Sources#edit

Marsden, Jerrold E.; Weinstein, Alan (1981). "Chapter 8: Inverse Functions and the Chain Rule". Calculus unlimited (PDF). Menlo Park, Calif.: Benjamin/Cummings Pub. Co. ISBN 0-8053-6932-5.

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