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:
[math]\displaystyle{ {\color{CornflowerBlue}{f'}}(x) = \frac{1}{{\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x))} }[/math]

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

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 [math]\displaystyle{ f }[/math] is denoted as [math]\displaystyle{ f^{-1} }[/math], where [math]\displaystyle{ f^{-1}(y) = x }[/math] if and only if [math]\displaystyle{ f(x) = y }[/math], then the inverse function rule is, in Lagrange's notation,

[math]\displaystyle{ \left[f^{-1}\right]'(a)=\frac{1}{f'\left( f^{-1}(a) \right)} }[/math].

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

[math]\displaystyle{ \mathcal{D}\left[f^{-1}\right]=\frac{1}{(\mathcal{D} f)\circ \left(f^{-1}\right)}, }[/math]

where [math]\displaystyle{ \mathcal{D} }[/math] denotes the unary derivative operator (on the space of functions) and [math]\displaystyle{ \circ }[/math] denotes function composition.

Geometrically, a function and inverse function have graphs that are reflections, in the line [math]\displaystyle{ y=x }[/math]. This reflection operation turns the gradient of any line into its reciprocal.^[1]

Assuming that [math]\displaystyle{ f }[/math] has an inverse in a neighbourhood of [math]\displaystyle{ x }[/math] and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at [math]\displaystyle{ 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]\displaystyle{ \frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = 1. }[/math]

This relation is obtained by differentiating the equation [math]\displaystyle{ f^{-1}(y)=x }[/math] in terms of $x$ and applying the chain rule, yielding that:

[math]\displaystyle{ \frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = \frac{dx}{dx} }[/math]

considering that the derivative of $x$ with respect to $x$ is 1.

Derivation

Let [math]\displaystyle{ f }[/math] be an invertible (bijective) function, let [math]\displaystyle{ x }[/math] be in the domain of [math]\displaystyle{ f }[/math], and let [math]\displaystyle{ y }[/math] be in the codomain of [math]\displaystyle{ f }[/math]. Since f is a bijective function, [math]\displaystyle{ y }[/math] is in the range of [math]\displaystyle{ f }[/math]. This also means that [math]\displaystyle{ y }[/math] is in the domain of [math]\displaystyle{ f^{-1} }[/math], and that [math]\displaystyle{ x }[/math] is in the codomain of [math]\displaystyle{ f^{-1} }[/math]. Since [math]\displaystyle{ f }[/math] is an invertible function, we know that [math]\displaystyle{ f(f^{-1}(y)) = y }[/math]. The inverse function rule can be obtained by taking the derivative of this equation.

[math]\displaystyle{ \dfrac{\mathrm{d}}{\mathrm{d}y} f(f^{-1}(y)) = \dfrac{\mathrm{d}}{\mathrm{d}y} y }[/math]

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

[math]\displaystyle{ \begin{align} \dfrac{\mathrm{d}\left( f(f^{-1}(y)) \right)}{\mathrm{d}\left( f^{-1}(y) \right)} \dfrac{\mathrm{d}\left(f^{-1}(y)\right)}{\mathrm{d}y} &= 1 \\ \dfrac{\mathrm{d}f(f^{-1}(y))}{\mathrm{d}f^{-1}(y)} \dfrac{\mathrm{d}f^{-1}(y)}{\mathrm{d}y} &= 1 \\ f^{\prime}(f^{-1}(y)) (f^{-1})^{\prime}(y) &= 1 \end{align} }[/math]

Rearranging then gives

[math]\displaystyle{ (f^{-1})^{\prime}(y) = \frac{1}{f^{\prime}(f^{-1}(y))} }[/math]

Rather than using [math]\displaystyle{ y }[/math] as the variable, we can rewrite this equation using [math]\displaystyle{ a }[/math] as the input for [math]\displaystyle{ f^{-1} }[/math], and we get the following:^[2]

[math]\displaystyle{ (f^{-1})^{\prime}(a) = \frac{1}{f^{\prime}\left( f^{-1}(a) \right)} }[/math]

Examples

[math]\displaystyle{ y = x^2 }[/math] (for positive $x$ ) has inverse [math]\displaystyle{ x = \sqrt{y} }[/math].

[math]\displaystyle{ \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]\displaystyle{ \frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = 2x \cdot\frac{1}{2x} = 1. }[/math]

At [math]\displaystyle{ 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]\displaystyle{ y = e^x }[/math] (for real $x$ ) has inverse [math]\displaystyle{ x = \ln{y} }[/math] (for positive [math]\displaystyle{ y }[/math])

[math]\displaystyle{ \frac{dy}{dx} = e^x \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{dx}{dy} = \frac{1}{y} = e^{-x} }[/math]

[math]\displaystyle{ \frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = e^x \cdot e^{-x} = 1. }[/math]

Additional properties

Integrating this relationship gives

[math]\displaystyle{ {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]\displaystyle{ 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]\displaystyle{ \int f^{-1}(x)\, {dx} = x f^{-1}(x) - F(f^{-1}(x)) + C }[/math]

Where [math]\displaystyle{ F }[/math] denotes the antiderivative of [math]\displaystyle{ 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]\displaystyle{ z = f'(x) }[/math] then we have, assuming [math]\displaystyle{ f''(x) \neq 0 }[/math]:[math]\displaystyle{ \frac{d(f')^{-1}(z)}{dz} = \frac{1}{f''(x)} }[/math]This can be shown using the previous notation [math]\displaystyle{ y = f(x) }[/math]. Then we have:

[math]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ n \ge 1 }[/math], with [math]\displaystyle{ z = f^{(n)}(x) }[/math], the nth derivative of f(x), and [math]\displaystyle{ y = f^{(n-1)}(x) }[/math], assuming [math]\displaystyle{ f^{(i)}(x) \neq 0 \text{ for } 0 \lt i \le n+1 }[/math]:

[math]\displaystyle{ \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]\displaystyle{ f^{-1}(f(x))=x }[/math] with respect to $x$ . One can continue the same process for higher derivatives. Differentiating the identity twice with respect to $x$ , one obtains

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \frac{d^2y}{dx^2} = - \frac{d^2x}{dy^2}\,\cdot\,\left(\frac{dy}{dx}\right)^3. }[/math]

Similarly for the third derivative:

[math]\displaystyle{ \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]\displaystyle{ \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 $f$ and $g$ are inverses, then

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

Example

[math]\displaystyle{ y = e^x }[/math] has the inverse [math]\displaystyle{ x = \ln y }[/math]. Using the formula for the second derivative of the inverse function,

[math]\displaystyle{ \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]\displaystyle{ \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.

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.

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Original source: https://en.wikipedia.org/wiki/Inverse function rule. Read more

[1] "Derivatives of Inverse Functions". https://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/inverseDeriv.html.

[2] "Derivatives of inverse functions". https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-2-new/ab-3-3/v/derivatives-of-inverse-functions.

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