Inverse function rule

Examples

• ${\displaystyle y=x^2}$ (for positive x) 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 real x) has inverse ${\displaystyle x=\ln y}$ (for positive ${\displaystyle y}$)

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

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

Additional properties

• 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 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:

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

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 the Legendre 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 ${\displaystyle f^{(i)}(x)\ne 0\text{ for }0<i\le n+1}$:

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

Higher order derivatives

The chain rule given above is obtained by differentiating the identity ${\displaystyle f^{-1}(y)=x}$ 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

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

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

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

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:

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

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:

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

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

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

• ${\displaystyle y=e^x}$ 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.