Difference between revisions of "Derivatives of Inverse Functions"

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[[File:Umkehrregel 2.png|thumb|right|250px|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>]]
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In mathematics, the '''inverse''' of a function <math>y = f(x)</math> is a function that, in some fashion, "undoes" the effect of <math>f</math>. 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>.
 
In mathematics, the '''inverse''' of a function <math>y = f(x)</math> is a function that, in some fashion, "undoes" the effect of <math>f</math>. 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>.
  
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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.
 
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.
  
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==Examples==
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* <math>y = x^2</math> (for positive {{Mvar|x}}) has inverse <math>x = \sqrt{y}</math>.
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:<math> \frac{dy}{dx} = 2x
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\mbox{ }\mbox{ }\mbox{ }\mbox{ };
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\mbox{ }\mbox{ }\mbox{ }\mbox{ }
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\frac{dx}{dy} = \frac{1}{2\sqrt{y}}=\frac{1}{2x} </math>
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:<math>\frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = 2x \cdot\frac{1}{2x}  =  1.</math>
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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.
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* <math>y = e^x</math> (for real {{Mvar|x}}) has inverse <math>x = \ln{y}</math> (for positive <math>y</math>)
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:<math> \frac{dy}{dx} = e^x
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\mbox{ }\mbox{ }\mbox{ }\mbox{ };
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\mbox{ }\mbox{ }\mbox{ }\mbox{ }
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\frac{dx}{dy} = \frac{1}{y} </math>
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:<math> \frac{dy}{dx}\,\cdot\,\frac{dx}{dy}  =  e^x \cdot \frac{1}{y}  =  \frac{e^x}{e^x}  =  1 </math>
  
  
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* [https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/03%3A_Derivatives/3.7%3A_Derivatives_of_Inverse_Functions Derivatives of Inverse Function], Mathematics LibreTexts
 
* [https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/03%3A_Derivatives/3.7%3A_Derivatives_of_Inverse_Functions Derivatives of Inverse Function], Mathematics LibreTexts
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==Licensing==
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Content obtained and/or adapted from:
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* [https://en.wikipedia.org/wiki/Inverse_functions_and_differentiation Inverse functions and differentiation, Wikipedia] under a CC BY-SA license

Latest revision as of 09:31, 28 October 2021

Rule:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\color{CornflowerBlue}{f'}}(x) = \frac{1}{{\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x))}}

Example for arbitrary :

In mathematics, the inverse of a function is a function that, in some fashion, "undoes" the effect of . The inverse of is denoted as , where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{-1}(y) = x} if and only if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = y} .

Their two derivatives, assuming they exist, are reciprocal, as the Leibniz notation suggests; that is:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = 1.}

This relation is obtained by differentiating the equation Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{-1}(y)=x} in terms of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} and applying the chain rule, yielding that:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = \frac{dx}{dx}}

considering that the derivative of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} is 1.

Writing explicitly the dependence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} on Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} , and the point at which the differentiation takes place, the formula for the derivative of the inverse becomes (in Lagrange's notation):

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[f^{-1}\right]'(a)=\frac{1}{f'\left( f^{-1}(a) \right)}} .

This formula holds in general whenever Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is continuous and injective on an interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} , with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} being differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{-1}(a)} (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \in I} ) and where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(f^{-1}(a)) \ne 0} . The same formula is also equivalent to the expression

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{D}\left[f^{-1}\right]=\frac{1}{(\mathcal{D} f)\circ \left(f^{-1}\right)},}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{D}} denotes the unary derivative operator (on the space of functions) and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \circ} denotes function composition.

Geometrically, a function and inverse function have graphs that are reflections, in the line Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=x} . This reflection operation turns the gradient of any line into its reciprocal.

Assuming that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} has an inverse in a neighbourhood of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} and have a derivative given by the above formula.

Examples

  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = x^2} (for positive x) has inverse Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \sqrt{y}} .
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx} = 2x \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{dx}{dy} = \frac{1}{2\sqrt{y}}=\frac{1}{2x} }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = 2x \cdot\frac{1}{2x} = 1.}

At Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.

  • Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y = e^x} (for real x) has inverse Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \ln{y}} (for positive Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y} )
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx} = e^x \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{dx}{dy} = \frac{1}{y} }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = e^x \cdot \frac{1}{y} = \frac{e^x}{e^x} = 1 }


Resources

Licensing

Content obtained and/or adapted from:

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