Inverse function rule

This article is about the computation of the derivative of an invertible function. For a condition on which a function is invertible, see Inverse function theorem.

The thick blue curve and the thick red curve are inverse to each other. A thin curve is the derivative of the same colored thick curve. Inverse function rule:
${\color {CornflowerBlue}{f'}}(x)={\frac {1}{{\color {Salmon}{(f^{-1})'}}({\color {Blue}{f}}(x))}}$

Example for arbitrary $x_{0}\approx 5.8$:
${\color {CornflowerBlue}{f'}}(x_{0})={\frac {1}{4}}$
${\color {Salmon}{(f^{-1})'}}({\color {Blue}{f}}(x_{0}))=4~$

In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable functionf 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,

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

This formula holds in general whenever $f$ is continuous and injective on an interval I, with $f$ being differentiable at $f^{-1}(y)$($\in I$) and where$f'(f^{-1}(y))\neq 0$. 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 $y=f(x).$ Let $g=f^{-1}.$ So, $f(g(y))=y.$ Derivating this equation with respect to ⁠$y$⁠, and using the chain rule, one gets

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

That is,

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

or

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

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 $y$)

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

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

Additional properties

• Integrating this relationship gives

${f^{-1}}(x)=\int {\frac {1}{f'({f^{-1}}(x))}}\,{dx}+C.$

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

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

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.

See also

• Calculus – Branch of mathematics
• Chain rule – For derivatives of composed functions
• Differentiation of trigonometric functions – Mathematical process of finding the derivative of a trigonometric function
• Differentiation rules – Rules for computing derivatives of functions
• Implicit function theorem – On converting relations to functions of several real variables
• Integration of inverse functions – Mathematical theorem, used in calculusPages displaying short descriptions of redirect targets
• Inverse function – Mathematical concept
• Inverse function theorem – Theorem in mathematics
• Table of derivatives – Rules for computing derivatives of functionsPages displaying short descriptions of redirect targets
• Vector calculus identities – Mathematical identities

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.