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}}

[[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:
${\color{CornflowerBlue}{f'}}(x) = \frac{1}{{\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x))}$

Example for arbitrary $x_0 \approx 5.8$ :
${\color{CornflowerBlue}{f'}}(x_0) = \frac{1}{4}$
${\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x_0)) = 4~$ ]]

In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function {{Mvar|f}} in terms of the derivative of {{Mvar|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,

: $\left[f^{-1}\right]'(y)=\frac{1}{f'\left( f^{-1}(y) \right)}$ .

This formula holds in general whenever $f$ is continuous and injective on an interval {{Mvar|I}}, with $f$ being differentiable at $f^{-1}(y)$ ( $\in I$ ) and where $f'(f^{-1}(y)) \ne 0$ . The same formula is also equivalent to the expression

: $\mathcal{D}\left[f^{-1}\right]=\frac{1}{(\mathcal{D} f)\circ \left(f^{-1}\right)},$

where $\mathcal{D}$ 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.{{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}}

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 {{Mvar|x}} and applying the chain rule, yielding that:

: $\frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = \frac{dx}{dx}$

considering that the derivative of {{Mvar|x}} with respect to {{Mvar|x}} is 1.

Derivation

Let $f$ be an invertible (bijective) function, let $x$ be in the domain of $f$ , and let $y=f(x).$ Let $g=f^{-1}.$ So, $f(g(y))=y.$ Derivating this equation with respect to {{tmath|y}}, and using the chain rule, one gets

: $f'(g(y))\cdot g'(y)=1.$

That is,

: $g'(y)=\frac 1 {f'(g(y))}$

: $(f^{-1})^{\prime}(y) = \frac{1}{f^{\prime}(f^{-1}(y))}.$

Examples

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

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

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

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 {{Mvar|x}}) has inverse $x = \ln{y}$ (for positive $y$ )

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

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

Additional properties

Integrating this relationship gives

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

:This is only useful if the integral exists. In particular we need $f'(x)$ 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 $z = f'(x)$ then we have, assuming $f$ (x) \neq 0: $\frac{d(f')^{-1}(z)}{dz} = \frac{1}{f$ (x)}This can be shown using the previous notation $y = f(x)$ . Then we have:

: $f'(x) = \frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} = \frac{dy}{dz} f$ (x) \Rightarrow \frac{dy}{dz} = \frac{f'(x) }{f(x)}Therefore:

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

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

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

Higher derivatives

The chain rule given above is obtained by differentiating the identity $f^{-1}(f(x))=x$ with respect to {{Mvar|x}}. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to {{Mvar|x}}, one obtains

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

that is simplified further by the chain rule as

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

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

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

Similarly for the third derivative:

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

or using the formula for the second derivative,

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

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

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

Example

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

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

so that

: $\frac{d^2x}{dy^2}\,\cdot\,y^3 + y = 0
\mbox{ }\mbox{ }\mbox{ }\mbox{ };
\mbox{ }\mbox{ }\mbox{ }\mbox{ }
\frac{d^2x}{dy^2} = -\frac{1}{y^2}$ ,

which agrees with the direct calculation.

References

{{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}}

Category:Articles containing proofs

Category:Differentiation rules

Category:Inverse functions

Category:Theorems in mathematical analysis

Category:Theorems in calculus