{{Short description|Formula for the derivative of an inverse function}} {{about|the computation of the derivative of an invertible function|a condition on which a function is invertible|Inverse function theorem}} {{refimprove|date=January 2022}} [[File:Umkehrregel 2.png|thumb|right|250px|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:<br><math>{\color{CornflowerBlue}{f'}}(x) = \frac{1}{{\color{Salmon}{\left[f^{-1}\right]'}}({\color{Blue}{f}}(x))}</math><br><br>Example for arbitrary <math>x_0 \approx 5.8</math>:<br><math>{\color{CornflowerBlue}{f'}}(x_0) = \frac{1}{4}</math><br><math>{\color{Salmon}{\left[f^{-1}\right]'}}({\color{Blue}{f}}(x_0)) = 4~</math>]] {{calculus|expanded=differential}} In calculus, the '''inverse function rule''' is a formula that expresses the derivative of the inverse of a bijective and differentiable function {{Mvar|f}} in terms of the derivative of {{Mvar|f}}. More precisely, if the inverse of <math>f</math> is denoted as <math>f^{-1}</math>, where <math>f^{-1}(y) = x</math> if and only if <math>f(x) = y</math>, then the inverse function rule is, in Lagrange's notation,
:<math>\left[f^{-1}\right]'(y)=\frac{1}{f'\left( f^{-1}(y) \right)}.</math>
This formula holds in general whenever <math>f</math> is continuous and injective on an interval {{Mvar|I}}, with <math>f</math> being differentiable at <math>f^{-1}(y)</math>(<math>\in I</math>) and where<math>f'(f^{-1}(y)) \ne 0</math>. The same formula is also equivalent to the expression
:<math>\mathcal{D}\left[f^{-1}\right]=\frac{1}{(\mathcal{D} f)\circ \left(f^{-1}\right)},</math>
where <math>\mathcal{D}</math> denotes the unary derivative operator (on the space of functions) and <math>\circ</math> denotes function composition.
Geometrically, a function and inverse function have graphs that are reflections, in the line <math>y=x</math>. This reflection operation turns the gradient of any line into its reciprocal.<ref>{{Cite web|url=https://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/inverseDeriv.html|title=Derivatives of Inverse Functions|website=oregonstate.edu|access-date=2019-07-26 |archive-url=https://web.archive.org/web/20210410154136/https://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/inverseDeriv.html |archive-date=2021-04-10 |url-status=dead}}</ref>
Assuming that <math>f</math> has an inverse in a neighbourhood of <math>x</math> and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at <math>x</math> 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,
:<math>\frac{dx}{dy}\,\frac{dy}{dx} = 1.</math>
This relation is obtained by differentiating the equation <math>f^{-1}(y)=x</math> in terms of {{Mvar|x}} and applying the chain rule, yielding that:
:<math>\frac{dx}{dy}\,\frac{dy}{dx} = \frac{dx}{dx}</math>
considering that the derivative of {{Mvar|x}} with respect to ''{{Mvar|x}}'' is 1.
== Derivation == Let <math>f</math> be an invertible (bijective) function, let <math>x</math> be in the domain of <math>f</math>, and let <math>y=f(x).</math> Let <math>g=f^{-1}.</math> So, <math>f(g(y))=y.</math> Differentiating this equation with respect to {{tmath|y}}, and using the chain rule, one gets :<math>f'(g(y))\cdot g'(y)=1.</math> That is, :<math>g'(y)=\frac 1 {f'(g(y))}</math> or :<math> \left[f^{-1}\right]^{\prime}(y) = \frac{1}{f^{\prime}(f^{-1}(y))}. </math>
== Examples ==
* <math>y = x^2</math> (for positive {{Mvar|x}}) has inverse <math>x = \sqrt{y}</math>.
:<math> \frac{dy}{dx} = 2x \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{dx}{dy} = \frac{1}{2\sqrt{y}}=\frac{1}{2x} </math>
:<math> \frac{dy}{dx}\,\frac{dx}{dy} = 2x \cdot\frac{1}{2x} = 1. </math>
At <math>x=0</math>, however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.
* <math>y = e^x</math> (for real {{Mvar|x}}) has inverse <math>x = \ln{y}</math> (for positive <math>y</math>)
:<math> \frac{dy}{dx} = e^x \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{dx}{dy} = \frac{1}{y} = e^{-x} </math>
:<math> \frac{dy}{dx}\,\frac{dx}{dy} = e^x e^{-x} = 1. </math>
==Additional properties==
* Integrating this relationship gives
::<math>{f^{-1}}(y)=\int\frac{1}{f'({f^{-1}}(y))}\,{dy} + C.</math>
:This is only useful if the integral exists. In particular we need <math>f'(x)</math> 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:
::<math> \int f^{-1}(y)\, {dy} = y f^{-1}(y) - F(f^{-1}(y)) + C </math>
:where <math> F </math> denotes the antiderivative of <math> f </math>.
* 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 <math> z = f'(x)</math> then we have, assuming <math> f''(x) \neq 0</math>:<math display="block"> \frac{d}{dz}\left[f'\right]^{-1}(z) = \frac{1}{f''(x)}</math>This can be shown using the previous notation <math> y = f(x)</math>. Then we have:
:<math display="block"> f'(x) = \frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} = \frac{dy}{dz} f''(x) \Rightarrow \frac{dy}{dz} = \frac{f'(x) }{f''(x)}</math>Therefore:
:<math> \frac{d}{dz}[f']^{-1}(z) = \frac{dx}{dz} = \frac{dy}{dz}\frac{dx}{dy} = \frac{f'(x)}{f''(x)}\frac{1}{f'(x)} = \frac{1}{f''(x)}</math>
By induction, we can generalize this result for any integer <math> n \ge 1</math>, with <math> z = f^{(n)}(x)</math>, the nth derivative of f(x), and <math> y = f^{(n-1)}(x)</math>, assuming <math> f^{(i)}(x) \neq 0 \text{ for } 0 < i \le n+1 </math>:
:<math> \frac{d}{dz}\left[f^{(n)}\right]^{-1}(z) = \frac{1}{f^{(n+1)}(x)}</math>
== Higher order derivatives ==
The chain rule given above is obtained by differentiating the identity <math>f^{-1}(y)=x</math> with respect to {{Mvar|y}}, where <math>y=f(x)</math>. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to ''{{Mvar|x}}'', one obtains
:<math> \frac{d^2y}{dx^2}\,\frac{dx}{dy} + \frac{d}{dx} \left(\frac{dx}{dy}\right)\,\left(\frac{dy}{dx}\right) = 0, </math>
that is simplified further by the chain rule as
:<math> \frac{d^2y}{dx^2}\,\frac{dx}{dy} + \frac{d^2x}{dy^2}\,\left(\frac{dy}{dx}\right)^2 = 0.</math>
Replacing the first derivative, using the identity obtained earlier, we get
:<math> \frac{d^2y}{dx^2} = - \frac{d^2x}{dy^2}\,\left(\frac{dy}{dx}\right)^3 </math>
which implies
:<math> \frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{\left(dy/dx\right)^3}. </math>
Similarly for the third derivative we have
:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\left(\frac{dy}{dx}\right)^4 - 3 \frac{d^2x}{dy^2}\,\frac{d^2y}{dx^2}\,\left(\frac{dy}{dx}\right)^2.</math>
Using the formula for the second derivative, we get
:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\left(\frac{dy}{dx}\right)^4 + 3 \left(\frac{d^2y}{dx^2}\right)^2\,\left(\frac{dy}{dx}\right)^{-1}</math>
which implies
:<math> \frac{d^3x}{dy^3} = - \frac{d^3y/dx^3}{\left(dy/dx\right)^4} + 3\frac{\left(d^2y/dx^2\right)^2}{\left(dy/dx\right)^5}. </math>
These formulas can also be written using Lagrange's notation:
:<math> \left[f^{-1}\right]''(y) = -\frac{f''(f^{-1}(y))}{\left[f'(f^{-1}(y))\right]^3}, </math>
:<math> \left[f^{-1}\right]'''(y) = -\frac{f'''(f^{-1}(y))}{\left[f'(f^{-1}(y))\right]^4} + 3\frac{\left[f''(f^{-1}(y))\right]^2}{\left[f'(f^{-1}(y))\right]^5}. </math>
In general, higher order derivatives of an inverse function can be expressed with Faà di Bruno's formula. Alternatively, the {{Mvar|n}}th derivative can be written succinctly as:
:<math> \left[f^{-1}\right]^{(n)}(y) = \left[\left(\frac{1}{f'(t)}\frac{d}{dt}\right)^{n} t\right]_{t = f^{-1}(y)}. </math>
From this expression, one can also derive the {{Mvar|n}}th-integration of inverse function with base-point {{Mvar|a}} using Cauchy formula for repeated integration whenever <math>f(f^{-1}(y)) = y</math>:
:<math> \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). </math>
=== Example ===
* <math>y = e^x</math> has the inverse <math>x = \ln y</math>. Using the formula for the second derivative of the inverse function,
:<math> \frac{dy}{dx} = \frac{d^2y}{dx^2} = e^x = y \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \left(\frac{dy}{dx}\right)^3 = y^3;</math>
so that
:<math> \frac{d^2x}{dy^2}\,\cdot\,y^3 + y = 0 \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{d^2x}{dy^2} = -\frac{1}{y^2},</math>
which agrees with the direct calculation.
== See also == {{Portal|Mathematics}}
* {{annotated link|Calculus}} * {{annotated link|Chain rule}} * {{annotated link|Differentiation of trigonometric functions}} * {{annotated link|Differentiation rules}} * {{annotated link|Implicit function theorem}} * {{annotated link|Integration of inverse functions}} * {{annotated link|Inverse function}} * {{annotated link|Inverse function theorem}} * {{annotated link|Table of derivatives}} * {{annotated link|Vector calculus identities}}
== References == {{Reflist}} * {{Cite book|last=Marsden|first=Jerrold E.|url=https://authors.library.caltech.edu/25054/10/CalcUch8-invfunc-chainrule.pdf|title=Calculus unlimited|date=1981|publisher=Benjamin/Cummings Pub. Co|first2=Alan |last2=Weinstein|isbn=0-8053-6932-5|location=Menlo Park, Calif.|chapter=Chapter 8: Inverse Functions and the Chain Rule}}
{{Calculus topics}}
Category:Articles containing proofs Category:Differentiation rules Category:Inverse functions Category:Theorems in mathematical analysis Category:Theorems in calculus