Inverse function rule explained

In calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function in terms of the derivative of . 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) = (1)/(f'\left( f ^{-1} (a) \right)) .

This formula holds in general whenever

f is continuous and injective on an interval, with f being differentiable at f ^{-1} (a) (\in I ) and wheref'(f ^{-1} (a)) \ne 0 . The same formula is also equivalent to the expressionl{D}\left[f ^{-1} \right] = (1)/(( l{D) f)\circ \left(f ^{-1} \right)},

where

l{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.^[1]

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,

(dx)/(dy) ⋅ (dy)/(dx) = 1 .

This relation is obtained by differentiating the equation

f ^{-1} (y) = x in terms of and applying the chain rule, yielding that:(dx)/(dy) ⋅ (dy)/(dx) = (dx)/(dx)

considering that the derivative of with respect to 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{ d } f(f^(y)) = \dfrac y

The right side is equal to 1 and the chain rule can be applied to the left side:

\begin{align} \dfrac{ d \left( f(f ^{-1} (y)) \right)}{ d \left( f ^{-1} (y) \right)} \dfrac{ d \left(f ^{-1} (y)\right)}{ d y} & = 1 \\ \dfrac{ d f(f ^{-1} (y))}{ d f ^{-1} (y)} \dfrac{ d f ^{-1} (y)}{ d y} & = 1 \\ f ^{\prime} (f ^{-1} (y)) (f ^{-1} ) ^{\prime} (y) & = 1 \end{align}

Rearranging then gives

(f ^{-1} ) ^{\prime} (y) = (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:^[2] (f ^{-1} ) ^{\prime} (a) = (1)/(f ^{\prime} \left( f ^{-1} (a) \right))

Examples

y = x ^{2} (for positive) has inverse x = \sqrt{y} .(dy)/(dx) = 2 x ; (dx)/(dy) = (1)/(2 \sqrt{y) }=\frac (dy)/(⋅ dx) (dx)/(dy) = 2 x ⋅ (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) has inverse x = ln {y} (for positive y )(dy)/(dx) = e ^{x ; (dx)/(dy)} = (1)/(y) = e ^{-x} (dy)/(⋅ dx) (dx)/(dy) = e ^{x} ⋅ e ^{-x} = 1 .

Additional properties

• Integrating this relationship gives

{f ^{-1} }(x)=\int\frac\, + 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} = x f ^{-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) ≠ 0 :$$\backslash frac\; =\; \backslash frac$$This can be shown using the previous notation y = f(x) . Then we have:

$$f\text{'}(x)\; =\; \backslash frac\; =\; \backslash frac\; \backslash frac\; =\; \backslash frac\; f$$(x) \Rightarrow \frac = \fracTherefore:

(d(f') ^{-1} (z))/(dz) = (dx)/(dz) = (dy)/(dz) (dx)/(dy) = (f'(x))/(f''(x)) (1)/(f'(x)) = (1)/(f''(x))

By induction, we can generalize this result for any integer

n \ge 1 , with z = f ^{(n)} (x) , the nth derivative of f(x), and y = f ^{(n-1)} (x) , assuming f ^{(i)} (x) ≠ 0 for 0 < i \le n +1 :(d(f ^{(n)} ) ^{-1} (z))/(dz) = (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 . One can continue the same process for higher derivatives. Differentiating the identity twice with respect to , one obtains(d ^{2y})/(⋅ dx ^{2}) (dx)/(dy) + (d)/(\left( dx) (dx)/(\right) ⋅ \left( dy) (dy)/(dx) \right) = 0 ,

that is simplified further by the chain rule as

(d ^{2y})/(⋅ dx ^{2}) (dx)/(dy) + (d ^{2x})/(⋅ \left( dy ^{2}) (dy)/(dx) \right) ^{2} = 0 .

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

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

Similarly for the third derivative:

(d ^{3y})/(dx ^{3}) = - (d ^{3x})/(⋅ \left( dy ^{3}) (dy)/(dx) \right) ^{4} - 3 (d ^{2x})/(⋅ dy ^{2}) (d ^{2y})/(⋅ \left( dx ^{2}) (dy)/(dx) \right) ^{2}

or using the formula for the second derivative,

(d ^{3y})/(dx ^{3}) = - (d ^{3x})/(⋅ \left( dy ^{3}) (dy)/(dx) \right) ^{4} + 3 \left( (d ^{2x})/(dy ^{2}) \right) ^{2 ⋅ \left( (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 and are inverses, then

g''(x) = (- 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,(dy)/(dx) = (d ^{2y})/(dx ^{2}) = e ^{x} = y ; \left( (dy)/(dx) \right) ^{3} = y ^{3;}

so that

(d ^{2x})/(dy ^{2}) ⋅ y ^{3} + y =0 ; (d ^{2x})/(dy ^{2}) =- (1)/(y ^{2}) ,

which agrees with the direct calculation.

References

• Book: Marsden, Jerrold E.. Calculus unlimited. 1981. Benjamin/Cummings Pub. Co. Alan . Weinstein. 0-8053-6932-5. Menlo Park, Calif.. Chapter 8: Inverse Functions and the Chain Rule.

Notes and References

1. Web site: Derivatives of Inverse Functions. oregonstate.edu. 2019-07-26 . https://web.archive.org/web/20210410154136/https://oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/inverseDeriv.html . 2021-04-10 . dead.
2. Web site: Derivatives of inverse functions . Khan Academy . 23 April 2022.