Inverse function rule

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,

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

This formula holds in general whenever ${\displaystyle f}$ is continuous and injective on an interval I, with ${\displaystyle f}$ being differentiable at ${\displaystyle f^{-1}(y)}$(${\displaystyle \in I}$) and where${\displaystyle f'(f^{-1}(y))\neq 0}$. 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,

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

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

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

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

Let ${\displaystyle f}$ be an invertible (bijective) function, let ${\displaystyle x}$ be in the domain of ${\displaystyle f}$, and let ${\displaystyle y=f(x).}$ Let ${\displaystyle g=f^{-1}.}$ So, ${\displaystyle f(g(y))=y.}$ Differentiating this equation with respect to ⁠${\displaystyle y}$⁠, and using the chain rule, one gets

${\displaystyle f'(g(y))\cdot g'(y)=1.}$

That is,

${\displaystyle g'(y)={\frac {1}{f'(g(y))}}}$

or

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

• ${\displaystyle y=x^{2}}$ (for positive x) has inverse ${\displaystyle x={\sqrt {y}}}$.

${\displaystyle {\frac {dy}{dx}}=2x{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{2{\sqrt {y}}}}={\frac {1}{2x}}}$

${\displaystyle {\frac {dy}{dx}}\,{\frac {dx}{dy}}=2x\cdot {\frac {1}{2x}}=1.}$

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 ${\displaystyle y}$)

${\displaystyle {\frac {dy}{dx}}=e^{x}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{y}}=e^{-x}}$

${\displaystyle {\frac {dy}{dx}}\,{\frac {dx}{dy}}=e^{x}e^{-x}=1.}$

• Integrating this relationship gives

${\displaystyle {f^{-1}}(y)=\int {\frac {1}{f'({f^{-1}}(y))}}\,{dy}+C.}$

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:

${\displaystyle \int f^{-1}(y)\,{dy}=yf^{-1}(y)-F(f^{-1}(y))+C}$

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 ${\displaystyle f''(x)\neq 0}$:$${\displaystyle {\frac {d}{dz}}\left[f'\right]^{-1}(z)={\frac {1}{f''(x)}}}$$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:

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

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 ${\displaystyle f^{(i)}(x)\neq 0{\text{ for }}0<i\leq n+1}$:

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

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

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

that is simplified further by the chain rule as

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

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

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

which implies

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

Similarly for the third derivative we have

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

Using the formula for the second derivative, we get

${\displaystyle {\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\left({\frac {dy}{dx}}\right)^{4}+3\left({\frac {d^{2}y}{dx^{2}}}\right)^{2}\,\left({\frac {dy}{dx}}\right)^{-1}}$

which implies

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

These formulas can also be written using Lagrange's notation:

${\displaystyle \left[f^{-1}\right]''(y)=-{\frac {f''(f^{-1}(y))}{\left[f'(f^{-1}(y))\right]^{3}}},}$

${\displaystyle \left[f^{-1}\right]'''(y)=-{\frac {f'''(f^{-1}(y))}{\left[f'(f^{-1}(y))\right]^{4}}}+3{\frac {\left[f''(f^{-1}(y))\right]^{2}}{\left[f'(f^{-1}(y))\right]^{5}}}.}$

In general, higher order derivatives of an inverse function can be expressed with Faà di Bruno's formula. Alternatively, the nth derivative can be written succinctly as:

${\displaystyle \left[f^{-1}\right]^{(n)}(y)=\left[\left({\frac {1}{f'(t)}}{\frac {d}{dt}}\right)^{n}t\right]_{t=f^{-1}(y)}.}$

From this expression, one can also derive the nth-integration of inverse function with base-point a using Cauchy formula for repeated integration whenever ${\displaystyle f(f^{-1}(y))=y}$:

${\displaystyle \left[f^{-1}\right]^{(-n)}(y)={\frac {1}{n!}}\left(f^{-1}(a)(y-a)^{n}+\int _{f^{-1}(a)}^{f^{-1}(y)}\left(y-f(u)\right)^{n}\,du\right).}$

• ${\displaystyle y=e^{x}}$ 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.