Inverse function rule

Short description: Calculus identity

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

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

In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function 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,

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

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$f^{\prime }(f^{-1}(a))\ne 0$. 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,

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

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.

$\frac{\mathrm{d}}{\mathrm{d}y}f(f^{-1}(y))=\frac{\mathrm{d}}{\mathrm{d}y}y$

The right side is equal to 1 and the chain rule can be applied to the left side:

$\begin{array}{rl}\frac{\mathrm{d}(f(f^{-1}(y)))}{\mathrm{d}(f^{-1}(y))}\frac{\mathrm{d}(f^{-1}(y))}{\mathrm{d}y}& =1\\ \frac{\mathrm{d}f(f^{-1}(y))}{\mathrm{d}{f}^{-1}(y)}\frac{\mathrm{d}{f}^{-1}(y)}{\mathrm{d}y}& =1\\ f^{\prime }(f^{-1}(y))(f^{-1}{)}^{\prime }(y)& =1\end{array}$

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:

$(f^{-1}{)}^{\prime }(a)=\frac{1}{f^{\prime }(f^{-1}(a))}$

Examples

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

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

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

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 $y$)

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

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

Additional properties

• Integrating this relationship gives

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

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:

$\int f^{-1}(x)dx=x{f}^{-1}(x)-F(f^{-1}(x))+C$

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 $z=f^{\prime }(x)$ then we have, assuming $f^{″}(x)\ne 0$:$$\frac{d(f^{\prime }{)}^{-1}(z)}{dz}=\frac{1}{f^{″}(x)}$$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:

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

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

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

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

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

that is simplified further by the chain rule as

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

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

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

Similarly for the third derivative:

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

or using the formula for the second derivative,

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

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

$g^{″}(x)=\frac{-f^{″}(g(x))}{[f^{\prime }(g(x)){]}^3}$

Example

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

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

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

• Calculus – Branch of mathematics
• Chain rule – For derivatives of composed functions
• 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
• Inverse function – Mathematical concept
• Inverse function theorem – Theorem in mathematics
• Vector calculus identities – Mathematical identities

References

• Marsden, Jerrold E.; Weinstein, Alan (1981). "Chapter 8: Inverse Functions and the Chain Rule". Calculus unlimited. Menlo Park, Calif.: Benjamin/Cummings Pub. Co. ISBN 0-8053-6932-5. https://authors.library.caltech.edu/25054/10/CalcUch8-invfunc-chainrule.pdf.