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Inverse function rule

## Summary

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 ${\displaystyle f}$ is denoted as ${\displaystyle f^{-1}}$, where ${\displaystyle f^{-1}(y)=x}$ if and only if ${\displaystyle f(x)=y}$, then the inverse function rule is, in Lagrange's notation,

${\displaystyle \left[f^{-1}\right]'(a)={\frac {1}{f'\left(f^{-1}(a)\right)}}}$.

This formula holds in general whenever ${\displaystyle f}$ is continuous and injective on an interval I, with ${\displaystyle f}$ being differentiable at ${\displaystyle f^{-1}(a)}$(${\displaystyle \in I}$) and where${\displaystyle f'(f^{-1}(a))\neq 0}$. The same formula is also equivalent to the expression

${\displaystyle {\mathcal {D}}\left[f^{-1}\right]={\frac {1}{({\mathcal {D}}f)\circ \left(f^{-1}\right)}},}$

where ${\displaystyle {\mathcal {D}}}$ denotes the unary derivative operator (on the space of functions) and ${\displaystyle \circ }$ denotes function composition.

Geometrically, a function and inverse function have graphs that are reflections, in the line ${\displaystyle y=x}$. This reflection operation turns the gradient of any line into its reciprocal.[1]

Assuming that ${\displaystyle f}$ has an inverse in a neighbourhood of ${\displaystyle x}$ and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at ${\displaystyle x}$ 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,

${\displaystyle {\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}=1.}$

This relation is obtained by differentiating the equation ${\displaystyle f^{-1}(y)=x}$ in terms of x and applying the chain rule, yielding that:

${\displaystyle {\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}={\frac {dx}{dx}}}$

considering that the derivative of x with respect to x is 1.

## Derivation

Let ${\displaystyle f}$  be an invertible (bijective) function, let ${\displaystyle x}$  be in the domain of ${\displaystyle f}$ , and let ${\displaystyle y}$  be in the codomain of ${\displaystyle f}$ . Since f is a bijective function, ${\displaystyle y}$  is in the range of ${\displaystyle f}$ . This also means that ${\displaystyle y}$  is in the domain of ${\displaystyle f^{-1}}$ , and that ${\displaystyle x}$  is in the codomain of ${\displaystyle f^{-1}}$ . Since ${\displaystyle f}$  is an invertible function, we know that ${\displaystyle f(f^{-1}(y))=y}$ . The inverse function rule can be obtained by taking the derivative of this equation.

${\displaystyle {\dfrac {\mathrm {d} }{\mathrm {d} y}}f(f^{-1}(y))={\dfrac {\mathrm {d} }{\mathrm {d} y}}y}$

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}}}

Rearranging then gives

${\displaystyle (f^{-1})^{\prime }(y)={\frac {1}{f^{\prime }(f^{-1}(y))}}}$

Rather than using ${\displaystyle y}$  as the variable, we can rewrite this equation using ${\displaystyle a}$  as the input for ${\displaystyle f^{-1}}$ , and we get the following:[2]

${\displaystyle (f^{-1})^{\prime }(a)={\frac {1}{f^{\prime }\left(f^{-1}(a)\right)}}}$

## Examples

• ${\displaystyle y=x^{2}}$  (for positive x) has inverse ${\displaystyle x={\sqrt {y}}}$ .
${\displaystyle {\frac {dy}{dx}}=2x{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{2{\sqrt {y}}}}={\frac {1}{2x}}}$
${\displaystyle {\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=2x\cdot {\frac {1}{2x}}=1.}$

At ${\displaystyle x=0}$ , however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.

• ${\displaystyle y=e^{x}}$  (for real x) has inverse ${\displaystyle x=\ln {y}}$  (for positive ${\displaystyle y}$ )
${\displaystyle {\frac {dy}{dx}}=e^{x}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{y}}=e^{-x}}$
${\displaystyle {\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=e^{x}\cdot e^{-x}=1.}$

${\displaystyle {f^{-1}}(x)=\int {\frac {1}{f'({f^{-1}}(x))}}\,{dx}+C.}$
This is only useful if the integral exists. In particular we need ${\displaystyle f'(x)}$  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:
${\displaystyle \int f^{-1}(x)\,{dx}=xf^{-1}(x)-F(f^{-1}(x))+C}$
Where ${\displaystyle F}$  denotes the antiderivative of ${\displaystyle f}$ .
• 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 ${\displaystyle z=f'(x)}$  then we have, assuming ${\displaystyle f''(x)\neq 0}$ :${\displaystyle {\frac {d(f')^{-1}(z)}{dz}}={\frac {1}{f''(x)}}}$ This can be shown using the previous notation ${\displaystyle y=f(x)}$ . Then we have:

${\displaystyle f'(x)={\frac {dy}{dx}}={\frac {dy}{dz}}{\frac {dz}{dx}}={\frac {dy}{dz}}f''(x)\Rightarrow {\frac {dy}{dz}}={\frac {f'(x)}{f''(x)}}}$ Therefore:
${\displaystyle {\frac {d(f')^{-1}(z)}{dz}}={\frac {dx}{dz}}={\frac {dy}{dz}}{\frac {dx}{dy}}={\frac {f'(x)}{f''(x)}}{\frac {1}{f'(x)}}={\frac {1}{f''(x)}}}$

By induction, we can generalize this result for any integer ${\displaystyle n\geq 1}$ , with ${\displaystyle z=f^{(n)}(x)}$ , the nth derivative of f(x), and ${\displaystyle y=f^{(n-1)}(x)}$ , assuming ${\displaystyle f^{(i)}(x)\neq 0{\text{ for }}0 :

${\displaystyle {\frac {d(f^{(n)})^{-1}(z)}{dz}}={\frac {1}{f^{(n+1)}(x)}}}$

## Higher derivatives

The chain rule given above is obtained by differentiating the identity ${\displaystyle f^{-1}(f(x))=x}$  with respect to x. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to x, one obtains

${\displaystyle {\frac {d^{2}y}{dx^{2}}}\,\cdot \,{\frac {dx}{dy}}+{\frac {d}{dx}}\left({\frac {dx}{dy}}\right)\,\cdot \,\left({\frac {dy}{dx}}\right)=0,}$

that is simplified further by the chain rule as

${\displaystyle {\frac {d^{2}y}{dx^{2}}}\,\cdot \,{\frac {dx}{dy}}+{\frac {d^{2}x}{dy^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{2}=0.}$

Replacing the first derivative, using the identity obtained earlier, we get

${\displaystyle {\frac {d^{2}y}{dx^{2}}}=-{\frac {d^{2}x}{dy^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{3}.}$

Similarly for the third derivative:

${\displaystyle {\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{4}-3{\frac {d^{2}x}{dy^{2}}}\,\cdot \,{\frac {d^{2}y}{dx^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{2}}$

or using the formula for the second derivative,

${\displaystyle {\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{4}+3\left({\frac {d^{2}x}{dy^{2}}}\right)^{2}\,\cdot \,\left({\frac {dy}{dx}}\right)^{5}}$

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

${\displaystyle g''(x)={\frac {-f''(g(x))}{[f'(g(x))]^{3}}}}$

## Example

• ${\displaystyle y=e^{x}}$  has the inverse ${\displaystyle x=\ln y}$ . Using the formula for the second derivative of the inverse function,
${\displaystyle {\frac {dy}{dx}}={\frac {d^{2}y}{dx^{2}}}=e^{x}=y{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}\left({\frac {dy}{dx}}\right)^{3}=y^{3};}$

so that

${\displaystyle {\frac {d^{2}x}{dy^{2}}}\,\cdot \,y^{3}+y=0{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {d^{2}x}{dy^{2}}}=-{\frac {1}{y^{2}}}}$ ,

which agrees with the direct calculation.