Inverse function theorem

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 continuous in an open interval containing $a$, differentiable at $a$ and $b=f(a)$. Suppose further that the derivative $f'(a)$ is nonzero, i.e., $f'(a)\neq 0$. 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 $f$ is continuous around the point, and $f'(f^{-1}(x))$ 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 $f$ is continuous around the point and $f'(f^{-1}(x))$ exists and is nonzero.
Pure Leibniz notation using dependent and independent variables	Suppose $y$ is a variable functionally dependent on $x$. Then, ${\frac {dy}{dx}}={\frac {1}{\frac {dx}{dy}}}$ (with domain caveats as above).

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 behavior of original function $f$ at $a$	Short version	Long version (using specific point, named functions)
increasing function from left	left hand derivative of $f^{-1}$ is related to left hand derivative of $f$	Suppose $f$ is an increasing function from the left at a point $a$. Suppose $b=f(a)$. Suppose further that the left hand derivative $f'_{-}(a)$ is nonzero, i.e., $f'_{-}(a)\neq 0$. Then the inverse function $f^{-1}$ is left differentiable at $b$, and further: $(f^{-1})'_{-}(b)={\frac {1}{f'_{-}(a)}}$
increasing function from right	right hand derivative of $f^{-1}$ is related to right hand derivative of $f$	Suppose $f$ is an increasing function from the right at a point $a$ (in other words, $f$ increases on the immediate right of $a$). Suppose $b=f(a)$. Suppose further that the right hand derivative $f'_{+}(a)$ is nonzero, i.e., $f'_{+}(a)\neq 0$. Then the inverse function $f^{-1}$ is right differentiable at $b$, and further: $(f^{-1})'_{+}(b)={\frac {1}{f'_{+}(a)}}$
decreasing function from left	right hand derivative of $f^{-1}$ is related to left hand derivative of $f$	Suppose $f$ is a decreasing function on the left at a point $a$. Suppose $b=f(a)$. Suppose further that the left hand derivative $f'_{-}(a)$ is nonzero, i.e., $f'_{-}(a)\neq 0$. Then the inverse function $f^{-1}$ is right differentiable at $b$, and further: $(f^{-1})'_{+}(b)={\frac {1}{f'_{-}(a)}}$
decreasing function from right	left hand derivative of $f^{-1}$ is related to right hand derivative of $f$	Suppose $f$ is a decreasing function from the right at a point $a$. Suppose $b=f(a)$. Suppose further that the right hand derivative $f'_{+}(a)$ is nonzero, i.e., $f'_{+}(a)\neq 0$. Then the inverse function $f^{-1}$ is left differentiable at $b$, and further: $(f^{-1})'_{-}(b)={\frac {1}{f'_{+}(a)}}$

Some additional notes:

• For a point in the interior of the domain at which the function is continuous, being increasing on the immediate left forces the function to be increasing on the immediate right, and vice versa. Similarly for decreasing.
• More generally, a continuous one-one function on an interval must be either increasing through the interval or decreasing throughout the interval.

We have been more specific in our statements in the table above to allow for the possibility of piecewise defined functions with discontinuities as well as to tackle the issue of interval endpoints where only one-sided notions make sense.

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)	$f(x):=-x^{1/3}$, $f^{-1}(x):=-x^{3}$, $a=b=0$

Significance

Version type	Significance
specific point, named functions (two-sided, finite)	This tells us that if a one-one function $f$ is differentiable at a point with nonzero derivative, then $f^{-1}$ is differentiable at the image of that point under $f$.
specific point, named functions (two-sided, infinity-sensitive)	This tells us that if a one-one function $f$ is either differentiable at a point or has a vertical tangent, then $f^{-1}$ is either differentiable at the image of that point or has a vertical tangent. Moreover, we can pair the possibilities for $f$ with the possibilities for $f^{-1}$ using the theorem.
specific point, named functions (one-sided version)	This tells us that if a one-one function $f$ is one-sided differentiable at a point, then the inverse function is one-sided differentiable at the image point, where the side remains the same for an increasing function and gets switched for a decreasing function.
generic point, named functions (two-sided, finite)	This tells us that the inverse of a differentiable one-one function with nowhere zero derivative is also a differentiable one-one function.
generic point, named functions (two-sided, infinity-sensitive)	This tells us that the inverse of a one-one function that is differentiable or has a vertical tangent at each point is also a one-one function that is either differentiable or has a vertical tangent at each point.
generic point, named functions (one-sided, infinity-sensitive)	This tells us that the inverse of a one-one function that is one-sided differentiable or has a (one or two-sided) vertical tangent at each point is also a one-one function that is one-sided differentiable or has a (one or two-sided) vertical tangent at each point.

Note two important caveats:

• The differentiable of $f$ at $a$ gives us information about the differentiability of $f^{-1}$, not at $a$, but at $f(a)$.
• The reciprocation means we have to be careful about zero and infinity. Thus, the inverse of a differentiable one-one function need not be differentiable everywhere on its domain.

Computational feasibility significance

Version type	Significance
specific point, named functions	Consider a one-one function $f$. It is possible to compute $(f^{-1})'(b)$ if we know the value of $f'(a)$ where $a=f^{-1}(b)$.
specific point, named functions (second version)	Consider a one-one function $f$. It is possible to compute $(f^{-1})'(b)$ if we know the generic expression for $f'$ and the specific value $f^{-1}(b)$.
generic point, named functions	Consider a one-one function $f$. It is possible to find a generic expression for $(f^{-1})'(x)$ in terms of $f'$ and $f^{-1}$. Note: [SHOW MORE] $f^{-1}$ itself may not have an explicit expression even if $f$ does, unless we treat inversion itself as a valid building block for writing expressions.

Computational results significance

See the section #Infinity-sensitive versions for some of the basic computational results in this direction.

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 $\cos \theta \geq 0$ 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 $\exp$	all real numbers	reciprocal function $x\mapsto 1/x$	exponential function $\exp$	By the inverse function theorem, the derivative at $x$ is ${\frac {1}{\ln '(\exp(x))}}={\frac {1}{1/(\exp x)}}=\exp(x)$