Inverse function rule

File:Umkehrregel 2.png

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

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

{{#invoke:sidebar|collapsible | class = plainlist | titlestyle = padding-bottom:0.25em; | pretitle = Part of a series of articles about | title = Calculus | image = <math>\int_{a}^{b} f'(t) \, dt = f(b) - f(a)</math> | listtitlestyle = text-align:center; | liststyle = border-top:1px solid #aaa;padding-top:0.15em;border-bottom:1px solid #aaa; | expanded = differential | abovestyle = padding:0.15em 0.25em 0.3em;font-weight:normal; | above =

• Fundamental theorem

• Limits
• Continuity

• Rolle's theorem
• Mean value theorem
• Inverse function theorem

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}} In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function Template:Mvar in terms of the derivative of Template:Mvar. 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 Template:Mvar, 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>{{#invoke:citation/CS1|citation |CitationClass=web }}</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}\,\cdot\, \frac{dy}{dx} = 1.</math>

This relation is obtained by differentiating the equation <math>f^{-1}(y)=x</math> in terms of Template:Mvar and applying the chain rule, yielding that:

<math>\frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = \frac{dx}{dx}</math>

considering that the derivative of Template:Mvar with respect to Template:Mvar 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 Template:Tmath, 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 Template:Mvar) 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 Template:Mvar) 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

• 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 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}(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 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 Template:Mvar. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to Template:Mvar, 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 Template:Mvar and Template:Mvar 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

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References

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