Editing Inverse function rule

{{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}{(f^{-1})'}}({\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}{(f^{-1})'}}({\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 function|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 function|continuous]] and [[Injective function|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 [[graph of a function|graphs]] that are [[Reflection (mathematics)|reflection]]s, in the line <math>y=x</math>. This reflection operation turns the [[slope|gradient]] of any line into its [[Multiplicative inverse|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 [[neighborhood (mathematics)|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}\,\cdot\, \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}\,\cdot\, \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> Derivating 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>
 (f^{-1})^{\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}\,\cdot\,\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}\,\cdot\,\frac{dx}{dy} = e^x \cdot e^{-x} = 1. </math>

==Additional properties==

* [[Integral|Integrating]] this relationship gives

::<math>{f^{-1}}(x)=\int\frac{1}{f'({f^{-1}}(x))}\,{dx} + 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 function|continuous]] derivative has an inverse in a [[neighbourhood (mathematics)|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}(x)\, {dx} = x f^{-1}(x) - F(f^{-1}(x)) + 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 transformation|Legendre transform]]. 
Let  <math> z = f'(x)</math>  then we have, assuming <math> f''(x) \neq 0</math>:<math display="block"> \frac{d(f')^{-1}(z)}{dz} = \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(f')^{-1}(z)}{dz} = \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(f^{(n)})^{-1}(z)}{dz} = \frac{1}{f^{(n+1)}(x)}</math>

==Higher derivatives==

The [[chain rule]] given above is obtained by differentiating the identity <math>f^{-1}(f(x))=x</math> with respect to {{Mvar|x}}. 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}\,\cdot\,\frac{dx}{dy} + \frac{d}{dx} \left(\frac{dx}{dy}\right)\,\cdot\,\left(\frac{dy}{dx}\right)  =  0, </math>

that is simplified further by the chain rule as

:<math> \frac{d^2y}{dx^2}\,\cdot\,\frac{dx}{dy} + \frac{d^2x}{dy^2}\,\cdot\,\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}\,\cdot\,\left(\frac{dy}{dx}\right)^3. </math>

Similarly for the third derivative:

:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\cdot\,\left(\frac{dy}{dx}\right)^4 -
3 \frac{d^2x}{dy^2}\,\cdot\,\frac{d^2y}{dx^2}\,\cdot\,\left(\frac{dy}{dx}\right)^2</math>

or using the formula for the second derivative,

:<math> \frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\cdot\,\left(\frac{dy}{dx}\right)^4 +
3 \left(\frac{d^2x}{dy^2}\right)^2\,\cdot\,\left(\frac{dy}{dx}\right)^5</math>

These formulas are generalized by the [[Faà di Bruno's formula]].

These formulas can also be written using Lagrange's notation. If ''{{Mvar|f}}'' and ''{{Mvar|g}}'' are inverses, then

:<math> g''(x) = \frac{-f''(g(x))}{[f'(g(x))]^3}</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]]