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 $f$ 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,

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

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

𝒟 [f^{- 1}] = \frac{1}{(𝒟 f) \circ (f^{- 1})},

where $𝒟$ 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{d x}{d y} \frac{d y}{d x} = 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{d x}{d y} \frac{d y}{d x} = \frac{d x}{d x}

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, $f (g (y)) = y .$ Differentiating 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}]}^{'} (y) = \frac{1}{f^{'} (f^{- 1} (y))} .

Examples

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

\frac{d y}{d x} = 2 x; \frac{d x}{d y} = \frac{1}{2 \sqrt{y}} = \frac{1}{2 x}

\frac{d y}{d x} \frac{d x}{d y} = 2 x \cdot \frac{1}{2 x} = 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{d y}{d x} = e^{x}; \frac{d x}{d y} = \frac{1}{y} = e^{- x}

\frac{d y}{d x} \frac{d x}{d y} = e^{x} e^{- x} = 1.

Additional properties

Integrating this relationship gives

f^{- 1} (y) = \int \frac{1}{f^{'} (f^{- 1} (y))} d y + 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} (y) d y = y f^{- 1} (y) - F (f^{- 1} (y)) + 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 $f^{″} (x) \neq 0$ : $\frac{d}{d z} {[f^{'}]}^{- 1} (z) = \frac{1}{f^{″} (x)}$ This can be shown using the previous notation $y = f (x)$ . Then we have:

f^{'} (x) = \frac{d y}{d x} = \frac{d y}{d z} \frac{d z}{d x} = \frac{d y}{d z} f^{″} (x) \Rightarrow \frac{d y}{d z} = \frac{f^{'} (x)}{f^{″} (x)}

Therefore:

\frac{d}{d z} [f^{'}]^{- 1} (z) = \frac{d x}{d z} = \frac{d y}{d z} \frac{d x}{d y} = \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 $f^{(i)} (x) \neq 0 for 0 < i \leq n + 1$ :

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

Higher order derivatives

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

\frac{d^{2} y}{d x^{2}} \frac{d x}{d y} + \frac{d}{d x} (\frac{d x}{d y}) (\frac{d y}{d x}) = 0,

that is simplified further by the chain rule as

\frac{d^{2} y}{d x^{2}} \frac{d x}{d y} + \frac{d^{2} x}{d y^{2}} {(\frac{d y}{d x})}^{2} = 0.

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

\frac{d^{2} y}{d x^{2}} = - \frac{d^{2} x}{d y^{2}} {(\frac{d y}{d x})}^{3}

which implies

\frac{d^{2} x}{d y^{2}} = - \frac{d^{2} y / d x^{2}}{{(d y / d x)}^{3}} .

Similarly for the third derivative we have

\frac{d^{3} y}{d x^{3}} = - \frac{d^{3} x}{d y^{3}} {(\frac{d y}{d x})}^{4} - 3 \frac{d^{2} x}{d y^{2}} \frac{d^{2} y}{d x^{2}} {(\frac{d y}{d x})}^{2} .

Using the formula for the second derivative, we get

\frac{d^{3} y}{d x^{3}} = - \frac{d^{3} x}{d y^{3}} {(\frac{d y}{d x})}^{4} + 3 {(\frac{d^{2} y}{d x^{2}})}^{2} {(\frac{d y}{d x})}^{- 1}

which implies

\frac{d^{3} x}{d y^{3}} = - \frac{d^{3} y / d x^{3}}{{(d y / d x)}^{4}} + 3 \frac{{(d^{2} y / d x^{2})}^{2}}{{(d y / d x)}^{5}} .

In general, higher order derivatives of an inverse function can be expressed with Faà di Bruno's formula. Alternatively, the $n$ th derivative can be written succinctly as an application of a differential operator:

{[f^{- 1}]}^{(n)} (y) = {[{(\frac{1}{f^{'} (t)} \frac{d}{d t})}^{n} t] |}_{t = f^{- 1} (y)} .

In Lagrange's notation, the second and third derivatives from the examples above can be found as applying the operator and substituting $t = f^{- 1} (y)$ after:

{[f^{- 1}]}^{″} (y) = {[\frac{1}{f^{'} (t)} \frac{d}{d t} (\frac{1}{f^{'} (t)} \frac{d}{d t} t)] |}_{t = f^{- 1} (y)} = - \frac{f^{″} (f^{- 1} (y))}{{[f^{'} (f^{- 1} (y))]}^{3}},

{[f^{- 1}]}^{‴} (y) = {[\frac{1}{f^{'} (t)} \frac{d}{d t} (\frac{1}{f^{'} (t)} \frac{d}{d t} (\frac{1}{f^{'} (t)} \frac{d}{d t} t))] |}_{t = f^{- 1} (y)} = - \frac{f^{‴} (f^{- 1} (y))}{{[f^{'} (f^{- 1} (y))]}^{4}} + 3 \frac{{[f^{″} (f^{- 1} (y))]}^{2}}{{[f^{'} (f^{- 1} (y))]}^{5}} .

One can also derive the $n$ th-integration of inverse function with base-point $a$ using Cauchy formula for repeated integration whenever $f (f^{- 1} (y)) = y$ :

{[f^{- 1}]}^{(- n)} (y) = \frac{1}{n!} (f^{- 1} (a) (y - a)^{n} + \int_{f^{- 1} (a)}^{f^{- 1} (y)} {(y - f (u))}^{n} d u) .

Example

$y = e^{x}$ has the inverse $x = \ln y$ . Using the formula for the second derivative of the inverse function,

\frac{d y}{d x} = \frac{d^{2} y}{d x^{2}} = e^{x} = y; {(\frac{d y}{d x})}^{3} = y^{3};

so that

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

which agrees with the direct calculation.

References

Citations

↑ "Derivatives of Inverse Functions". oregonstate.edu. Archived from the original on 2021-04-10. Retrieved 2019-07-26.

Sources

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.

[1] "Derivatives of Inverse Functions". oregonstate.edu. Archived from the original on 2021-04-10. Retrieved 2019-07-26.