174.3.125.23: needs more links

2014-10-09T06:24:15Z

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2013-07-11T06:30:02Z

Dating maintenance tags: {{Single source}} {{Inline citations}}

← Older revision		Revision as of 08:30, 11 July 2013
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	~~I am Shad from Bakersfield~~. ~~I am learning to play~~ the ~~Trumpet~~. ~~Other hobbies~~ are ~~Seaglass collecting~~.<br><br>~~Have~~ a ~~look~~ at ~~my page~~: [~~http~~://~~x1b~~.it/~~workathomelegitimatejobs459492 when~~ are the ~~benefits~~ of ~~wake up~~ 2 ~~rank up magic~~ the ~~seventh one~~]		[[File:Umkehrregel 2.png\|thumb\|right\|250px\|Rule:<br><math>{\color{Periwinkle}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{Periwinkle}f'}(x_0) = \frac{1}{4}</math><br><math>{\color{Salmon}(f^{-1})'}({\color{Blue}f}(x_0)) = 4~</math>]]
			In [[mathematics]], the '''inverse''' of a [[function (mathematics)\|function]] <math>y = f(x)</math> is a function that, in some fashion, "undoes" the effect of <math>f</math> (see [[inverse function]] for a formal and detailed definition). The inverse of <math>f</math> is denoted <math>f^{-1}</math>. The statements {{nowrap\|1=''y'' = ''f''(''x'')}} and {{nowrap\|1=''x'' = ''f''<sup> −1</sup>(''y'')}} are equivalent.

			Their two derivatives, assuming they exist, are [[Reciprocal (mathematics)\|reciprocal]], as the [[Leibniz notation]] suggests; that is:

			:<math>\frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = 1. </math>

			This is a direct consequence of the [[chain rule]], since

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

			and the derivative of <math>x</math> with respect to <math>x</math> is 1.

			Writing explicitly the dependence of <math>y</math> on <math>x</math> and the point at which the differentiation takes place and using Lagrange's notation, the formula for the derivative of the inverse becomes

			:<math>\left[f^{-1}\right]'(a)=\frac{1}{f'\left( f^{-1}(a) \right)}</math>

			Geometrically, a function and inverse function have graphs that are [[Reflection (mathematics)\|reflection]]s, in the line {{nowrap\|1=''y'' = ''x''}}. This reflection operation turns the [[gradient]] of any line into its [[Multiplicative inverse\|reciprocal]].

			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.

			==Examples==

			* <math>\,y = x^2</math> (for positive <math>x</math>) 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 {{nowrap\|1=''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.

			* <math>\,y = e^x</math> (for real <math>x</math>) 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} </math>

			:<math> \frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = e^x \cdot \frac{1}{y} = \frac{e^x}{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 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.

			==Higher derivatives==

			The [[chain rule]] given above is obtained by differentiating the identity {{nowrap\|1=''x'' = ''f''<sup> −1</sup>(''f''(''x''))}} with respect to ''x''. One can continue the same process for higher derivatives. Twice differentiating the identity with respect to ''x ''obtains,

			:<math> \frac{d^2y}{dx^2}\,\cdot\,\frac{dx}{dy} + \frac{d^2x}{dy^2}\,\cdot\,\left(\frac{dy}{dx}\right)^2 = 0 </math>

			or replacing the first derivative using the formula above,

			:<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 ''f'' and ''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==

			*[[Calculus]]
			*[[Inverse function]]s
			*[[Chain rule]]
			*[[Inverse function theorem]]
			*[[Implicit function theorem]]
			*[[Integration of inverse functions]]

			{{DEFAULTSORT:Inverse Functions And Differentiation}}
			[[Category:Differentiation rules]]
			[[Category:Inverse functions]]

en>Dmcq: /* Generalizations */ remove unnecessary header

2012-06-27T23:58:05Z

Generalizations: remove unnecessary header

New page

I am Shad from Bakersfield. I am learning to play the Trumpet. Other hobbies are Seaglass collecting.<br><br>Have a look at my page: [http://x1b.it/workathomelegitimatejobs459492 when are the benefits of wake up 2 rank up magic the seventh one]

Power rule - Revision history

174.3.125.23: needs more links

en>AnomieBOT: Dating maintenance tags: {{Single source}} {{Inline citations}}

en>Dmcq: /* Generalizations */ remove unnecessary header