Inverse function rule

The inverse function rule is a fundamental theorem in calculus that expresses the derivative of the inverse of a bijective and differentiable function f in terms of the derivative of f itself, allowing computation without explicitly solving for the inverse. Specifically, if y = f(x) defines a one-to-one function with a differentiable inverse x = f^{-1}(y), then the derivative of the inverse function is \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}, or equivalently, if g = f^{-1}, then g'(a) = \frac{1}{f'(g(a))} for any a in the domain where f' is nonzero.^[1] This rule applies to functions whose inverses exist and are differentiable, typically requiring f to be continuously differentiable and strictly monotonic on its domain./03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions) The rule originates from the implicit differentiation of the composition f(g(x)) = x, which yields f'(g(x)) \cdot g'(x) = 1, directly leading to the formula after solving for g'(x). It is particularly useful in practical applications, such as finding the slope of inverse trigonometric functions like \arcsin(x) or \arctan(x), where explicit differentiation is straightforward but inverting the original function (e.g., \sin(y) = x) would otherwise be cumbersome. For instance, the derivative of \arcsin(x) is \frac{1}{\sqrt{1 - x^2}}, derived by applying the rule to f(y) = \sin(y).^[2] Beyond basic derivatives, the rule extends to higher-order derivatives and multivariable settings via the inverse function theorem, which guarantees local invertibility under certain conditions on the Jacobian./03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions) This theorem underscores the symmetry between a function and its inverse, highlighting how differentiation preserves structural relationships in calculus. It forms a cornerstone for more advanced topics, including optimization problems involving inverse relations and numerical methods for root-finding, where implicit functions arise frequently.^[1]

Statement of the Rule

Formal Statement

The inverse function of a function f: A \to B is a function f^{-1}: B \to A such that f(f^{-1}(y)) = y for all y in B and f^{-1}(f(x)) = x for all x in A; this requires f to be bijective, swapping the roles of domain and codomain between f and f^{-1}.^[1] In standard notation, if y = f(x), then the inverse relation is x = f^{-1}(y), or equivalently y = f^{-1}(x) when expressing the inverse explicitly.^[3] The inverse function rule provides the derivative of this inverse under suitable conditions. Specifically, if f is differentiable on an interval containing f^{-1}(x) and f'(f^{-1}(x)) \neq 0, then the derivative of the inverse is given by

(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))},

where f' denotes the derivative of f.^[1] This formula holds provided f is continuously differentiable or satisfies conditions ensuring the inverse is also differentiable./03:_Derivatives/3.07:_Derivatives_of_Inverse_Functions) This differentiation rule originates in calculus as a direct consequence of the chain rule applied to the composition of f and f^{-1}.^[1]

Assumptions and Conditions

The inverse function rule, which provides the derivative of the inverse of a differentiable function, requires that the original function f be bijective on its domain, meaning it is both one-to-one (injective) and onto (surjective), to ensure the existence of an inverse function f^{-1}.^[3] Without bijectivity, the inverse may not be well-defined globally, as multiple inputs could map to the same output or the range might not cover the required domain.^[4] For the derivative of the inverse to exist at a point x, the function f must be differentiable at the corresponding point f^{-1}(x), and the derivative f'(f^{-1}(x)) must be nonzero to prevent division by zero in the rule's expression.^[4] In many contexts, f is assumed to be continuously differentiable (i.e., f' is continuous) near the point of interest, which strengthens the guarantees for the inverse's smoothness.^[5] The rule applies locally where these conditions hold, even if f is not globally invertible; for instance, a function may require domain restrictions to achieve bijectivity in a neighborhood around the evaluation point.^[5] This local applicability aligns with the inverse function theorem, which states that if f is continuously differentiable at an interior point c with f'(c) \neq 0, then f is locally invertible near c and the local inverse is differentiable at f(c).^[5] These conditions collectively ensure the validity of the inverse function rule as stated formally.^[4]

Derivation

Derivation via Implicit Differentiation

To derive the inverse function rule using implicit differentiation, begin by letting y = f^{-1}(x), which implies the relation x = f(y).^[6] Differentiate both sides of x = f(y) with respect to x, treating y as a function of x. The left side yields \frac{dx}{dx} = 1, while the right side requires the chain rule: \frac{d}{dx} [f(y)] = f'(y) \cdot \frac{dy}{dx}. Thus, the equation becomes

$1 = f'(y) \cdot \frac{dy}{dx}.

^[7] Solving for \frac{dy}{dx}, divide both sides by f'(y), assuming f'(y) \neq 0 (consistent with the differentiability and local invertibility conditions for f):

\frac{dy}{dx} = \frac{1}{f'(y)}.

^[8] Substitute back y = f^{-1}(x) to express the derivative in terms of x:

\frac{d}{dx} [f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}. $$ This completes the algebraic derivation of the inverse function rule.[](https://web.ma.utexas.edu/users/m408n/CurrentWeb/LM3-5-5.php) To verify the steps, consider a simple [linear function](/page/Linear_function) $ f(y) = ay + b $ where $ a \neq 0 $. Here, $ x = ay + b $, so differentiating gives $ 1 = a \cdot \frac{dy}{dx} $, and solving yields $ \frac{dy}{dx} = \frac{1}{a} $. Substituting $ y = f^{-1}(x) = \frac{x - b}{a} $ into $ \frac{1}{f'(y)} = \frac{1}{a} $ confirms the result matches directly, illustrating the manipulation without additional complexity.[](https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/157e1a358da2e7815736937bf75b8942_MIT18_01SCF10_Ses15a.pdf) ### Geometric Interpretation The graph of an inverse function $f^{-1}$ is the reflection of the graph of $f$ across the line $y = x$, swapping the coordinates of corresponding points such that if $(a, b)$ lies on the graph of $f$, then $(b, a)$ lies on the graph of $f^{-1}$. This symmetry implies that tangent lines at corresponding points are also reflections across $y = x$, transforming their slopes into reciprocals.[](https://mathbooks.unl.edu/Calculus/sec-2-6-inverse.html) To interpret the slopes geometrically, consider a point $(y, x)$ on the graph of $f$, where $x = f(y)$; the slope of the tangent line to $f$ at this point is $f'(y)$. Upon reflection, this point maps to $(x, y)$ on the graph of $f^{-1}$, and the tangent slope at that point becomes $1 / f'(y)$, as the reflection operation inverts the rise-over-run ratio of the tangent line.[](https://mathbooks.unl.edu/Calculus/sec-2-6-inverse.html) This reciprocal relationship visually justifies the algebraic rule for the derivative of the inverse function. The condition $f'(y) \neq 0$ has a clear geometric meaning: a [horizontal](/page/Horizontal) tangent to $f$ (slope zero) at $(y, x)$ reflects to a [vertical tangent](/page/Vertical_tangent) (undefined slope) on $f^{-1}$ at $(x, y)$, where the [inverse function](/page/Inverse_function) fails to be [differentiable](/page/Differentiable_function).[](https://mathvault.ca/derivative-inverse-functions/) Thus, non-zero [derivative](/page/Derivative)s ensure that tangents to $f$ remain non-vertical after reflection, preserving differentiability for $f^{-1}$. This geometric framework connects to the [inverse function theorem](/page/Inverse_function_theorem), which guarantees that a continuously [differentiable function](/page/Differentiable_function) with non-zero [derivative](/page/Derivative) at a point is locally invertible in a neighborhood, as the non-zero [slope](/page/Slope) ensures the [graph](/page/Graph) is strictly increasing or decreasing, admitting a unique local inverse with the reciprocal [derivative](/page/Derivative) property.[](https://www.jirka.org/ra/html/sec_ift.html) ## Examples ### Basic Examples To illustrate the inverse function rule, consider simple cases where the [inverse](/page/Inverse) can be found explicitly and the [derivative](/page/Derivative) verified both directly and via the formula $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$, provided $f'(f^{-1}(x)) \neq 0$.[](https://openstax.org/books/calculus-volume-1/pages/3-7-derivatives-of-inverse-functions) **Example 1: [Linear function](/page/Linear_function)** Consider $f(x) = 2x + 1$. To find the [inverse](/page/Inverse), solve $y = 2x + 1$ for $x$: $x = \frac{y - 1}{2}$, so $f^{-1}(x) = \frac{x - 1}{2}$. Differentiating directly gives $(f^{-1})'(x) = \frac{1}{2}$. Now apply the rule: $f'(x) = 2$, so $f'(f^{-1}(x)) = 2 \neq 0$, and $(f^{-1})'(x) = \frac{1}{2}$, matching the direct result. This holds for all $x$ since $f'$ is [constant](/page/Constant) and nonzero.[](https://openstax.org/books/calculus-volume-1/pages/3-7-derivatives-of-inverse-functions) **Example 2: Cubic function** For $f(x) = x^3$, the inverse is found by solving $y = x^3$: $x = y^{1/3}$, so $f^{-1}(x) = x^{1/3}$. Direct differentiation yields $(f^{-1})'(x) = \frac{1}{3} x^{-2/3}$. Using the rule: $f'(x) = 3x^2$, so $f'(f^{-1}(x)) = 3 (x^{1/3})^2 = 3 x^{2/3} \neq 0$ for $x \neq 0$, and $(f^{-1})'(x) = \frac{1}{3 x^{2/3}}$, which agrees. At $x = 0$, the derivative is undefined, consistent with the condition.[](https://openstax.org/books/calculus-volume-1/pages/3-7-derivatives-of-inverse-functions) These examples demonstrate the rule's utility in computing derivatives without always solving for the explicit inverse, as long as the nonzero derivative condition is satisfied.[](https://openstax.org/books/calculus-volume-1/pages/3-7-derivatives-of-inverse-functions) ### Trigonometric and Exponential Examples The sine function, when restricted to the domain $[-\pi/2, \pi/2]$, is continuous and strictly increasing, ensuring it is bijective onto its range $[-1, 1]$.[](https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.07%3A_Derivatives_of_Inverse_Functions) This domain restriction is necessary to define a single-valued inverse function, known as the arcsine or inverse sine, denoted $\arcsin x$, with domain $x \in [-1, 1]$ and range $[-\pi/2, \pi/2]$.[](https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.07%3A_Derivatives_of_Inverse_Functions) To derive the derivative of $\arcsin x$ using the inverse function rule, let $y = \arcsin x$, so $x = \sin y$ where $y \in [-\pi/2, \pi/2]$. Differentiating both sides with respect to $x$ yields $1 = \cos y \cdot y'$, so $y' = 1 / \cos y$. Substituting $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$ (taking the positive root since $\cos y \geq 0$ in the interval) gives

\frac{d}{dx} [\arcsin x] = \frac{1}{\sqrt{1 - x^2}}.

[](https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.07%3A_Derivatives_of_Inverse_Functions)[](https://mathworld.wolfram.com/InverseSine.html) This formula holds for $x \in (-1, 1)$, where the expression is defined and the function is differentiable. Similarly, for the arctangent function, let $y = \arctan x$, so $x = \tan y$ where $y \in (-\pi/2, \pi/2)$. Differentiating both sides with respect to $x$ yields $1 = \sec^2 y \cdot y'$, so $y' = 1 / \sec^2 y = \cos^2 y$. Since $\sec^2 y = 1 + \tan^2 y = 1 + x^2$, it follows that

\frac{d}{dx} [\arctan x] = \frac{1}{1 + x^2}.

This holds for all $x \in \mathbb{R}$, as $\tan y$ is strictly increasing on $(-\pi/2, \pi/2)$ and its derivative is always positive.[](https://mathworld.wolfram.com/InverseTangent.html) For the exponential function $f(x) = e^x$, which is strictly increasing and bijective from $\mathbb{R}$ to $(0, \infty)$, the inverse is the natural logarithm $\ln x$, defined for $x > 0$ with [range](/page/Range) $\mathbb{R}$.[](https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.07%3A_Derivatives_of_Inverse_Functions) Applying the inverse function [rule](/page/Rule), let $y = \ln x$, so $x = e^y$. Differentiating gives $1 = e^y \cdot y'$, hence $y' = 1 / e^y = 1 / x$. Thus,

\frac{d}{dx} [\ln x] = \frac{1}{x},

for $x > 0$.[](https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.07%3A_Derivatives_of_Inverse_Functions)[](https://mathworld.wolfram.com/NaturalLogarithm.html) This matches the direct application of the [rule](/page/Rule), confirming the [derivative](/page/Derivative) without needing to compute limits or other methods. In cases where an explicit [inverse function](/page/Inverse_function) is known but its direct differentiation is cumbersome—such as involving trigonometric identities or limits—the [inverse function](/page/Inverse_function) rule offers a practical alternative by leveraging the [derivative](/page/Derivative) of the original [function](/page/Function) evaluated at the inverse point.[](https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.07%3A_Derivatives_of_Inverse_Functions) For trigonometric inverses like arcsine and arctangent, this approach avoids resolving complex expressions from the definition, while for exponentials, it simplifies verification across the [domain](/page/Domain). Such applications highlight the rule's utility in computational contexts, particularly under [domain](/page/Domain) restrictions that guarantee invertibility. ## Properties and Extensions ### Additional Properties One key property arising from the inverse function rule is the composition of a function with its inverse, which yields the [identity function](/page/Identity_function). Specifically, the derivative of the composition $ (f^{-1} \circ f)'(x) = [1](/page/1) $ and $ (f \circ f^{-1})'(x) = [1](/page/1) $, as these compositions differentiate to the constant [1](/page/1), consistent with the derivative of the [identity function](/page/Identity_function).[](https://math.montana.edu/burroughs/documents/InversesFunctions.pdf) The inverse function rule is intimately related to the chain rule, appearing as a special case where the outer function in a composition is the inverse of the inner function. By applying the chain rule to $ f(f^{-1}(x)) = x $, differentiation yields $ f'(f^{-1}(x)) \cdot (f^{-1})'(x) = [1](/page/1) $, directly implying the inverse derivative formula $ (f^{-1})'(x) = \frac{[1](/page/1)}{f'(f^{-1}(x))} $.[](https://math.montana.edu/burroughs/documents/InversesFunctions.pdf) This relation highlights the symmetry in the derivatives of inverse functions: if $ g = f^{-1} $, then $ g'(x) \cdot f'(g(x)) = 1 $, underscoring the reciprocal nature of their slopes at corresponding points. Geometrically, this reciprocity reflects the fact that the graphs of $ f $ and $ g $ are symmetric across the line $ y = x $, so the slope of one at a point is the reciprocal of the slope of the other at the reflected point.[](https://activecalculus.org/single/sec-2-6-inverse.html) At critical points of $ f $ where $ f'(y) = 0 $, the inverse function rule indicates that the derivative of the inverse becomes infinite, provided the inverse exists locally. This behavior manifests as a [vertical tangent](/page/Vertical_tangent) in the graph of the [inverse function](/page/Inverse_function), as seen in cases where the original function has a horizontal tangent.[](https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/03%3A_Derivatives/3.07%3A_Derivatives_of_Inverse_Functions) ### Higher-Order Derivatives The extension of the inverse function rule to higher-order derivatives allows for the computation of subsequent derivatives of the [inverse function](/page/Inverse_function) $f^{-1}$, assuming $f$ is sufficiently differentiable and $f'$ is nowhere zero in the relevant domain. For the second derivative, if $y = f^{-1}(x)$, then differentiating the first-order relation $ \frac{dy}{dx} = \frac{1}{f'(y)} $ with respect to $x$ yields

\frac{d^2 y}{dx^2} = -\frac{f''(y)}{[f'(y)]^3},

where the differentiation applies the [quotient rule](/page/Quotient_rule) and [chain rule](/page/Chain_rule): the derivative of the numerator is zero, while the denominator's derivative is $f''(y) \cdot \frac{dy}{dx}$, substituted back to obtain the expression.[](https://calculus.subwiki.org/wiki/Second_derivative_rule_for_inverse_function) For the general $n$th derivative, denoted $(f^{-1})^{(n)}(x)$, an explicit formula can be derived using [Faà di Bruno's formula](/page/Faà_di_Bruno's_formula) adapted to the composition arising from the inverse relation $f(f^{-1}(x)) = x$. This adaptation expresses the higher derivatives in terms of [Bell polynomials](/page/Bell_polynomials) or recursive sums involving the derivatives of $f$, often resulting in expressions of the form

(f^{-1})^{(n)}(x) = \sum \frac{(-1)^{k} n!}{[f'(f^{-1}(x))]^{n+1}} \cdot P_{n,k}(f''(f^{-1}(x)), \dots, f^{(n)}(f^{-1}(x))),

where $P_{n,k}$ are partial [Bell polynomials](/page/Bell_polynomials) capturing the partitions of $n$, though closed forms grow combinatorially complex for $n > 2$.[](https://mathoverflow.net/questions/98501/fa%C3%A0-di-brunos-formula-for-inverse-functions) A special case occurs for linear functions $f(x) = ax + b$ with $a \neq 0$, where the first [derivative](/page/Derivative) $f'(x) = a$ is constant, so $f''(x) = 0$ and all higher [derivatives](/page/Hartshorn) vanish; consequently, $(f^{-1})^{(n)}(x) = 0$ for all $n \geq 2$, consistent with the [inverse](/page/Inverse) $f^{-1}(x) = (x - b)/a$ being linear. Computing higher-order [derivatives](/page/Hartshorn) for $n > 2$ typically involves significant combinatorial [expansion](/page/Expansion), making [symbolic](/page/The_Symbolic) [computation](/page/Computation) tools such as Mathematica or [SymPy](/page/SymPy) essential for practical evaluation beyond low orders.[](https://vixra.org/pdf/1703.0295v1.pdf)