Derivative of Inverse Function

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Theorem

Let $I = \closedint a b$ and $J = \closedint c d$ be closed real intervals.

Let $I^o = \openint a b$ and $J^o = \openint c d$ be the corresponding open real intervals.


Let $f: I \to J$ be a real function which is continuous on $I$ and differentiable on $I^o$ such that $J = f \left({I}\right)$.

Let either:

$\forall x \in I^o: D \, \map f x > 0$

or:

$\forall x \in I^o: D \, \map f x < 0$


Then:

$f^{-1}: J \to I$ exists and is continuous on $J$
$f^{-1}$ is differentiable on $J^o$
$\forall y \in J^o: D \, \map {f^{-1} } y = \dfrac 1 {D \, \map f x}$


Proof

From Derivative of Monotone Function, it follows that $f$ is either:

strictly increasing on $I$ (if $\forall x \in I^o: D \, \map f x > 0$)

or:

strictly decreasing on $I$ (if $\forall x \in I^o: D \, \map f x < 0$).

Therefore from Inverse of Strictly Monotone Function‎ it follows that $f^{-1}: J \to I$ exists.

As $f$ is continuous, from Image of Interval by Continuous Function is Interval it follows that $J$ is an interval.

By the Corollary to Limit of Increasing Function and the Corollary to Limit of Decreasing Function, $f^{-1}: J \to I$ is continuous.


Next its derivative is to be considered.

Suppose $f$ is strictly increasing.

Let $y \in J^o$.

Then $\map {f^{-1} } y \in I^o$.

Let $k = \map {f^{-1} } {y + h} - \map {f^{-1} } y$.

Thus:

$\map {f^{-1} } {y + h} = \map {f^{-1} } y + k = x + k$

Thus:

$y + h = \map f {x + k}$

and hence:

$h = \map f {x + k} - y = \map f {x + k} - \map f x$

Since $f^{-1}$ is continuous on $J$, it follows that $k \to 0$ as $h \to 0$.

Also, $f^{-1}$ is strictly increasing from Inverse of Strictly Monotone Function‎ and so $k \ne 0$ unless $h = 0$.

So by Limit of Composite Function we get:

$\dfrac {\map {f^{-1} } {y + h} - \map {f^{-1} } y} h = \dfrac k {\map f {x + k} - \map f x}$

Thus:

$\dfrac {\map {f^{-1} } {y + h} - \map {f^{-1} } y} h \to \dfrac 1 {D \, \map f x}$

as $h \to 0$.


Suppose $f$ is strictly decreasing.

Exactly the same argument applies.

$\blacksquare$


Style Note

Leibniz's notation for derivatives $\paren {\dfrac {\d y} {\d x} }$ allows for a particularly elegant statement of this rule:

$\dfrac {\d x} {\d y} = \dfrac 1 {\frac {\d y} {\d x} }$

where:

$\dfrac {\d x} {\d y}$ is the derivative of $x$ with respect to $y$
$\dfrac {\d y} {\d x}$ is the derivative of $y$ with respect to $x$.

However, do not interpret this to mean that derivatives can be treated as fractions, it is simply a convenient notation.


Sources