Difference between revisions of "Inverse function theorem"

From Calculus
Jump to: navigation, search
(One-sided version)
Line 57: Line 57:
 
|-
 
|-
 
| undefined, but approaching <math>-\infty</math>, i.e., [[vertical tangent]] || zero || -- || Both decreasing. <math>f</math> is decreasing through <math>a</math> (with the rate of decrease peaking to <math>\infty</math>) and <math>f^{-1}</math> is decreasing through <math>b</math> (though the rate of decrease dips to zero because it's a [[point of inflection]] with horizontal tangent) || <math>f(x) := -x^{1/3}</math>, <math>f^{-1}(x) := -x^3</math>, <math>a = b = 0</math>
 
| undefined, but approaching <math>-\infty</math>, i.e., [[vertical tangent]] || zero || -- || Both decreasing. <math>f</math> is decreasing through <math>a</math> (with the rate of decrease peaking to <math>\infty</math>) and <math>f^{-1}</math> is decreasing through <math>b</math> (though the rate of decrease dips to zero because it's a [[point of inflection]] with horizontal tangent) || <math>f(x) := -x^{1/3}</math>, <math>f^{-1}(x) := -x^3</math>, <math>a = b = 0</math>
 +
|}
 +
 +
==Examples==
 +
 +
===Generic point examples===
 +
 +
Below we list some examples of functions and their inverse functions to which the inverse function theorem can be fruitfully applied.
 +
 +
{| class="sortable" border="1"
 +
! Original function !! Domain on which it restricts to a [[one-one function]] !! Inverse function for the restriction to that domain !! Domain of inverse function (equals range of original function) !! Derivative of original function !! Derivative of inverse function !! Explanation using inverse function theorem
 +
|-
 +
| [[sine function]] <math>\sin</math> || <math>[-\pi/2,\pi/2]</math> || [[arc sine function]] <math>\arcsin</math> || <math>[-1,1]</math> || [[cosine function]] <math>\cos</math> || <math>\frac{1}{\sqrt{1 - x^2}}</math> || By the inverse function theorem, the derivative at <math>x</math> is <math>\frac{1}{\sin'(\arcsin x)} = \frac{1}{\cos(\arcsin x)}</math>. Use that <math>\cos \theta \ge 0</math> on the range of <math>\arcsin</math> and <math>\cos^2\theta + \sin^2 \theta = 1</math> to get that <math>\cos(\arcsin x) = \sqrt{1 - x^2}</math>
 +
|-
 +
| [[tangent function]] <math>\tan</math> || <math>(-\pi/2,\pi/2)</math> || [[arc tangent function]] <math>\arctan</math> || all real numbers || [[secant-squared function]] <math>\sec^2</math> || <math>\frac{1}{1 + x^2}</math> || By the inverse function theorem, the derivative at <math>x</math> is <math>\frac{1}{\tan'(\arctan x)} = \frac{1}{\sec^2(\arctan x)}</math>. Use that <math>\sec^2 \theta = 1 + \tan^2\theta</math> and get <math>\sec^2(\arctan x) = 1 + x^2</math>.
 +
|-
 +
| [[natural logarithm]] <math>\ln</math> || <math>(0,\infty)</math> || [[exponential function]] <math>\exp</math> || [[reciprocal function]] <math>x \mapsto 1/x</math> || [[exponential function]] <math>\exp</math> || By the inverse function theorem, the derivative at <math>x</math> is <math>\frac{1}{\ln'(\exp(x))} = \frac{1}{1/(\exp x)} = \exp(x)</math>
 
|}
 
|}

Revision as of 17:41, 16 December 2011

This article is about a differentiation rule, i.e., a rule for differentiating a function expressed in terms of other functions whose derivatives are known.
View other differentiation rules

Statement

Verbal statement

The derivative of the inverse function at a point equals the reciprocal of the derivative of the function at its inverse image point.

Statement with symbols

Version type Statement
specific point, named functions Suppose f is a function of one variable that is a one-one function and a is in the domain of f. Suppose f is differentiable at a and b = f(a). Suppose further that the derivative f'(a) is nonzero, i.e., f'(a) \ne 0. Then the inverse function f^{-1} is differentiable at b, and further:
(f^{-1})'(b) = \frac{1}{f'(a)}
generic point, named functions, point notation Suppose f is a function of one variable that is a one-one function. Then, the formula for the derivative of the inverse function is as follows:
\! (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}
with the formula applicable at all points in the range of f for which f'(f^{-1}(x)) exists and is nonzero.
generic point, named functions, point-free notation Suppose f is a function of one variable that is a one-one function. Then, the formula for the derivative of the inverse function is as follows:
\! (f^{-1})'= \frac{1}{f' \circ f^{-1}}
with the formula applicable at all points in the range of f for which f'(f^{-1}(x)) exists and is nonzero.
Pure Leibniz notation using dependent and independent variables Suppose y is a variable functionally dependent on x. Then, \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}
MORE ON THE WAY THIS DEFINITION OR FACT IS PRESENTED: We first present the version that deals with a specific point (typically with a \{ \}_0 subscript) in the domain of the relevant functions, and then discuss the version that deals with a point that is free to move in the domain, by dropping the subscript. Why do we do this?
The purpose of the specific point version is to emphasize that the point is fixed for the duration of the definition, i.e., it does not move around while we are defining the construct or applying the fact. However, the definition or fact applies not just for a single point but for all points satisfying certain criteria, and thus we can get further interesting perspectives on it by varying the point we are considering. This is the purpose of the second, generic point version.

One-sided version

One-sided versions exist, but we need to be careful about issues of left and right. We state the two cases:

Case for function Short version Long version (using specific point, named functions)
increasing function left hand derivative of f^{-1} is related to left hand derivative of f Suppose f is an increasing function of one variable. Suppose b = f(a). Suppose further that the left hand derivative f'_-(a) is nonzero, i.e., f'_-(a) \ne 0. Then the inverse function f^{-1} is left differentiable at b, and further:
(f^{-1})'_-(b) = \frac{1}{f'_-(a)}
increasing function right hand derivative of f^{-1} is related to right hand derivative of f Suppose f is an increasing function of one variable. Suppose b = f(a). Suppose further that the right hand derivative f'_+(a) is nonzero, i.e., f'_+(a) \ne 0. Then the inverse function f^{-1} is right differentiable at b, and further:
(f^{-1})'_+(b) = \frac{1}{f'_+(a)}
decreasing function right hand derivative of f^{-1} is related to left hand derivative of f Suppose f is a decreasing function of one variable. Suppose b = f(a). Suppose further that the left hand derivative f'_-(a) is nonzero, i.e., f'_-(a) \ne 0. Then the inverse function f^{-1} is right differentiable at b, and further:
(f^{-1})'_+(b) = \frac{1}{f'_-(a)}
decreasing function left hand derivative of f^{-1} is related to right hand derivative of f Suppose f is a decreasing function of one variable. Suppose b = f(a). Suppose further that the right hand derivative f'_+(a) is nonzero, i.e., f'_+(a) \ne 0. Then the inverse function f^{-1} is left differentiable at b, and further:
(f^{-1})'_-(b) = \frac{1}{f'_+(a)}

Infinity-sensitive versions

The following version accounts for the infinity cases. We provide only the specific point, named functions version. Assume that f is a one-one function that is continuous at a point a in its domain, with b = f(a). There are six cases of interest:

Case for f'(a) Case for (f^{-1})'(b) Relation between them Increase/decrease? Example
undefined, but approaching +\infty zero -- Both increasing. f is increasing through a (with the rate of increase peaking to \infty) and f^{-1} is increasing through b (though the rate of increase dips to zero because it's a point of inflection with horizontal tangent). f(x) := x^{1/3}, f^{-1}(x) := x^3, a = b = 0
positive positive reciprocals of each other. Both increasing. f is increasing through a and f^{-1} is increasing through b. f(x) := x^3, f^{-1}(x) := x^{1/3}, a = b = 1
zero undefined, but approaching +\infty, i.e., vertical tangent -- Both increasing. f is increasing through a (though the rate of increase dips to zero because it's a point of inflection with horizontal tangent) and f^{-1} is increasing through b (with the rate of increase peaking to \infty). f(x) := x^3, f^{-1}(x) := x^{1/3}, a = b = 0
zero undefined, but approaching -\infty, i.e., vertical tangent -- Both decreasing. f is decreasing through a (though the rate of decrease dips to zero because it's a point of inflection with horizontal tangent) and f^{-1} is decreasing through b (with the rate of decrease peaking to \infty). f(x) := -x^3, f^{-1}(x) := -x^{1/3}, a = b = 0
negative negative reciprocals of each other Both decreasing. f is decreasing through a and f^{-1} is decreasing through b. f(x) := -x^3, f^{-1}(x) := -x^{1/3}, a = 1,b = -1
undefined, but approaching -\infty, i.e., vertical tangent zero -- Both decreasing. f is decreasing through a (with the rate of decrease peaking to \infty) and f^{-1} is decreasing through b (though the rate of decrease dips to zero because it's a point of inflection with horizontal tangent) f(x) := -x^{1/3}, f^{-1}(x) := -x^3, a = b = 0

Examples

Generic point examples

Below we list some examples of functions and their inverse functions to which the inverse function theorem can be fruitfully applied.

Original function Domain on which it restricts to a one-one function Inverse function for the restriction to that domain Domain of inverse function (equals range of original function) Derivative of original function Derivative of inverse function Explanation using inverse function theorem
sine function \sin [-\pi/2,\pi/2] arc sine function \arcsin [-1,1] cosine function \cos \frac{1}{\sqrt{1 - x^2}} By the inverse function theorem, the derivative at x is \frac{1}{\sin'(\arcsin x)} = \frac{1}{\cos(\arcsin x)}. Use that \cos \theta \ge 0 on the range of \arcsin and \cos^2\theta + \sin^2 \theta = 1 to get that \cos(\arcsin x) = \sqrt{1 - x^2}
tangent function \tan (-\pi/2,\pi/2) arc tangent function \arctan all real numbers secant-squared function \sec^2 \frac{1}{1 + x^2} By the inverse function theorem, the derivative at x is \frac{1}{\tan'(\arctan x)} = \frac{1}{\sec^2(\arctan x)}. Use that \sec^2 \theta = 1 + \tan^2\theta and get \sec^2(\arctan x) = 1 + x^2.
natural logarithm \ln (0,\infty) exponential function \exp reciprocal function x \mapsto 1/x exponential function \exp By the inverse function theorem, the derivative at x is \frac{1}{\ln'(\exp(x))} = \frac{1}{1/(\exp x)} = \exp(x)