Inverse function rule

File:Umkehrregel 2.png

The thick blue curve and the thick red curve are inverse to each other. A thin curve is the derivative of the same colored thick curve. Inverse function rule:
$f^{\prime }(x)=\frac{1}{\left[f^{-1}\right](f(x))}$

Example for arbitrary $x_0\approx 5.8$:
$f^{\prime }(x_0)=\frac{1}{4}$
$\left[f^{-1}\right](f(x_0))=4$

In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function Template:Mvar in terms of the derivative of Template:Mvar. 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,

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

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

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

where ${𝒟}$ 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,

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

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

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

considering that the derivative of Template:Mvar with respect to Template:Mvar is 1.

Derivation

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

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

That is,

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

or

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

Examples

• $y=x^2$ (for positive Template:Mvar) has inverse $x=\sqrt{y}$.

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

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

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 Template:Mvar) has inverse $x=\ln y$ (for positive $y$)

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

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

Additional properties

• Integrating this relationship gives

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

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

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

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

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

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

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

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

Higher order derivatives

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

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

that is simplified further by the chain rule as

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

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

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

which implies

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

Similarly for the third derivative we have

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

Using the formula for the second derivative, we get

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

which implies

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

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

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

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

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

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

From this expression, one can also derive the Template:Mvarth-integration of inverse function with base-point Template:Mvar using Cauchy formula for repeated integration whenever $f(f^{-1}(y))=y$:

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

Example

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

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

so that

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

which agrees with the direct calculation.

See also

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