This article needs additional citations for verification. Please help improve this article by adding citations to reliable sources. Unsourced material may be challenged and removed.Find sources: "Inverse function rule" – news · newspapers · books · scholar · JSTOR (January 2022) (Learn how and when to remove this template message) 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: f ′ ( x ) = 1 ( f − 1 ) ′ ( f ( x ) ) {\displaystyle {\color {CornflowerBlue}{f'))(x)={\frac {1}((\color {Salmon}{(f^{-1})'))({\color {Blue}{f))(x))))} Example for arbitrary x 0 ≈ 5.8 {\displaystyle x_{0}\approx 5.8} : f ′ ( x 0 ) = 1 4 {\displaystyle {\color {CornflowerBlue}{f'))(x_{0})={\frac {1}{4))} ( f − 1 ) ′ ( f ( x 0 ) ) = 4   {\displaystyle {\color {Salmon}{(f^{-1})'))({\color {Blue}{f))(x_{0}))=4~}

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

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

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

${\mathcal {D))\left[f^{-1}\right]={\frac {1}{({\mathcal {D))f)\circ \left(f^{-1}\right))),$ where ${\mathcal {D))$ denotes the unary derivative operator (on the space of functions) and $\circ$ denotes function composition.

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

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

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

${\frac {dx}{dy))\,\cdot \,{\frac {dy}{dx))={\frac {dx}{dx))$ considering that the derivative of x with respect to x is 1.

## Derivation

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

${\dfrac {\mathrm {d} }{\mathrm {d} y))f(f^{-1}(y))={\dfrac {\mathrm {d} }{\mathrm {d} y))y$ The right side is equal to 1 and the chain rule can be applied to the left side:

{\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

$(f^{-1})^{\prime }(y)={\frac {1}{f^{\prime }(f^{-1}(y))))$ Rather than using $y$ as the variable, we can rewrite this equation using $a$ as the input for $f^{-1)$ , and we get the following:

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

• $y=x^{2)$ (for positive x) has inverse $x={\sqrt {y))$ .
${\frac {dy}{dx))=2x{\mbox{ )){\mbox{ )){\mbox{ )){\mbox{ ));{\mbox{ )){\mbox{ )){\mbox{ )){\mbox{ )){\frac {dx}{dy))={\frac {1}{2{\sqrt {y))))={\frac {1}{2x))$ ${\frac {dy}{dx))\,\cdot \,{\frac {dx}{dy))=2x\cdot {\frac {1}{2x))=1.$ At $x=0$ , however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.

• $y=e^{x)$ (for real x) has inverse $x=\ln {y)$ (for positive $y$ )
${\frac {dy}{dx))=e^{x}{\mbox{ )){\mbox{ )){\mbox{ )){\mbox{ ));{\mbox{ )){\mbox{ )){\mbox{ )){\mbox{ )){\frac {dx}{dy))={\frac {1}{y))=e^{-x)$ ${\frac {dy}{dx))\,\cdot \,{\frac {dx}{dy))=e^{x}\cdot e^{-x}=1.$ ${f^{-1))(x)=\int {\frac {1}{f'({f^{-1))(x))))\,{dx}+C.$ This is only useful if the integral exists. In particular we need $f'(x)$ 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:
$\int f^{-1}(x)\,{dx}=xf^{-1}(x)-F(f^{-1}(x))+C$ Where $F$ denotes the antiderivative of $f$ .
• 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 $z=f'(x)$ then we have, assuming $f''(x)\neq 0$ :

${\frac {d(f')^{-1}(z)}{dz))={\frac {1}{f''(x)))$ This can be shown using the previous notation $y=f(x)$ . Then we have:

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

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

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

${\frac {d^{2}y}{dx^{2))}\,\cdot \,{\frac {dx}{dy))+{\frac {d}{dx))\left({\frac {dx}{dy))\right)\,\cdot \,\left({\frac {dy}{dx))\right)=0,$ that is simplified further by the chain rule as

${\frac {d^{2}y}{dx^{2))}\,\cdot \,{\frac {dx}{dy))+{\frac {d^{2}x}{dy^{2))}\,\cdot \,\left({\frac {dy}{dx))\right)^{2}=0.$ Replacing the first derivative, using the identity obtained earlier, we get

${\frac {d^{2}y}{dx^{2))}=-{\frac {d^{2}x}{dy^{2))}\,\cdot \,\left({\frac {dy}{dx))\right)^{3}.$ Similarly for the third derivative:

${\frac {d^{3}y}{dx^{3))}=-{\frac {d^{3}x}{dy^{3))}\,\cdot \,\left({\frac {dy}{dx))\right)^{4}-3{\frac {d^{2}x}{dy^{2))}\,\cdot \,{\frac {d^{2}y}{dx^{2))}\,\cdot \,\left({\frac {dy}{dx))\right)^{2)$ or using the formula for the second derivative,

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

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

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

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

which agrees with the direct calculation.