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Incalculus, theinverse function rule is aformula that expresses thederivative of theinverse of abijective anddifferentiable functionf in terms of the derivative off. 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, inLagrange's notation,

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

This formula holds in general whenever${\displaystyle f}$ iscontinuous andinjective on an intervalI, with${\displaystyle f}$ being differentiable at${\displaystyle f^{-1}(y)}$(${\displaystyle \in I}$) and where${\displaystyle f^{\prime }(f^{-1}(y))\ne 0}$. The same formula is also equivalent to the expression

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

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

Geometrically, a function and inverse function havegraphs that arereflections, in the line${\displaystyle y=x}$. This reflection operation turns thegradient of any line into itsreciprocal.^[1]

Assuming that${\displaystyle f}$ has an inverse in aneighbourhood 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 inLeibniz'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 ofx and applying thechain rule, yielding that:

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

considering that the derivative ofx with respect tox is 1.

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 thechain rule, one gets

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

That is,

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

or

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

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

${\displaystyle \frac{dy}{dx}=2x;\frac{dx}{dy}=\frac{1}{2\sqrt{y}}=\frac{1}{2x}}$

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

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 realx) has inverse${\displaystyle x=\ln y}$ (for positive${\displaystyle y}$)

${\displaystyle \frac{dy}{dx}=e^x;\frac{dx}{dy}=\frac{1}{y}=e^{-x}}$

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

• Integrating this relationship gives

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

This is only useful if the integral exists. In particular we need${\displaystyle f^{\prime }(x)}$ to be non-zero across the range of integration.

It follows that a function that has acontinuous derivative has an inverse in aneighbourhood 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:

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

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 theLegendre transform.

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

$${\displaystyle f^{\prime }(x)=\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{dy}{dz}{f}^{″}(x)\Rightarrow \frac{dy}{dz}=\frac{f^{\prime }(x)}{f^{″}(x)}}$$Therefore:

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

By induction, we can generalize this result for any integer${\displaystyle n\ge 1}$, with${\displaystyle z=f^{(n)}(x)}$, the nth derivative of f(x), and${\displaystyle y=f^{(n-1)}(x)}$, assuming:

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

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

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

that is simplified further by the chain rule as

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

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

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

which implies

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

Similarly for the third derivative we have

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

Using the formula for the second derivative, we get

${\displaystyle \frac{d^{3}y}{d{x}^3}=-\frac{d^{3}x}{d{y}^3}{\left(\frac{dy}{dx}\right)}^4+3{\left(\frac{d^{2}y}{d{x}^2}\right)}^2{\left(\frac{dy}{dx}\right)}^{-1}}$

which implies

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

These formulas can also be written using Lagrange's notation:

${\displaystyle {\left[f^{-1}\right]}^{″}(y)=-\frac{f^{″}(f^{-1}(y))}{{\left[f^{\prime }(f^{-1}(y))\right]}^3},}$

${\displaystyle {\left[f^{-1}\right]}^{‴}(y)=-\frac{f^{‴}(f^{-1}(y))}{{\left[f^{\prime }(f^{-1}(y))\right]}^4}+3\frac{{\left[f^{″}(f^{-1}(y))\right]}^2}{{\left[f^{\prime }(f^{-1}(y))\right]}^5}.}$

In general, higher order derivatives of an inverse function can be expressed withFaà di Bruno's formula. Alternatively, thenth derivative can be written succinctly as:

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

From this expression, one can also derive thenth-integration of inverse function with base-pointa usingCauchy formula for repeated integration whenever${\displaystyle f(f^{-1}(y))=y}$:

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

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

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

so that

${\displaystyle \frac{d^{2}x}{d{y}^2}\cdot y^3+y=0;\frac{d^{2}x}{d{y}^2}=-\frac{1}{y^2},}$

which agrees with the direct calculation.

• Calculus – Branch of mathematics
• Chain rule – Formula in calculus
• Differentiation of trigonometric functions – Mathematical process of finding the derivative of a trigonometric function
• Differentiation rules – Rules for computing derivatives of functions
• Implicit function theorem – On converting relations to functions of several real variables
• Integration of inverse functions – Mathematical theorem, used in calculusPages displaying short descriptions of redirect targets
• Inverse function – Mathematical concept
• Inverse function theorem – Theorem in mathematics
• Table of derivatives – Rules for computing derivatives of functionsPages displaying short descriptions of redirect targets
• Vector calculus identities – Mathematical identities

• 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.

• Differentiation rules
• Inverse functions
• Theorems in mathematical analysis
• Theorems in calculus

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