Derivative of Inverse Function

Theorem
Let $I = \left[{a .. b}\right]$ and $J = \left[{c .. d}\right]$ be closed real intervals.

Let $I^o = \left({a .. b}\right)$ and $J^o = \left({c .. d}\right)$ 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:
 * $\forall x \in I^o: D f \left({x}\right) > 0$, or:
 * $\forall x \in I^o: D f \left({x}\right) < 0$

Then:
 * $f^{-1}: J \to I$ exists and is continuous on $J$
 * $f^{-1}$ is differentiable on $J^o$
 * $\forall y \in J^o: D f^{-1} \left({y}\right) = \dfrac 1 {D f \left({x}\right)}$

Proof
From Derivative of Monotone Function, it follows that $f$ is either:
 * strictly increasing on $I$ (if $\forall x \in I^o: D f \left({x}\right) > 0$), or:
 * strictly decreasing on $I$ (if $\forall x \in I^o: D f \left({x}\right) < 0$).

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 it follows that $J$ is an interval.

By the corollary of Limit of Monotone Function, $f^{-1}: J \to I$ is continuous.

Now we consider its derivative.


 * Suppose $f$ is strictly increasing.

Let $y \in J^o$.

Then $f^{-1} \left({y}\right) \in I^o$.

Let $k = f^{-1} \left({y + h}\right) - f^{-1} \left({y}\right)$.

Thus:
 * $f^{-1} \left({y + h}\right) = f^{-1} \left({y}\right) + k = x + k$

Thus:
 * $y + h = f \left({x + k}\right)$

and hence:
 * $h = f \left({x + k}\right) - y = f \left({x + k}\right) - f \left({x}\right)$

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:


 * $\displaystyle \frac {f^{-1} \left({y + h}\right) - f^{-1} \left({y}\right)} {h} = \frac {k} {f \left({x + k}\right) - f \left({x}\right)}$

Thus:
 * $\displaystyle \frac {f^{-1} \left({y + h}\right) - f^{-1} \left({y}\right)} {h} \to \frac 1 {D f \left({x}\right)}$

as $h \to 0$.


 * Suppose $f$ is strictly decreasing.

Exactly the same argument applies.

Style Note
Leibniz's notation for derivatives $\left (\dfrac{\mathrm dy}{\mathrm dx}\right )$ allows for a particularly elegant statement of this rule:


 * $\displaystyle \frac{\mathrm dx}{\mathrm dy} = \frac 1 {\frac{\mathrm dy}{\mathrm dx}}$

where:
 * $\dfrac{\mathrm dx}{\mathrm dy}$ is the derivative of $x$ with respect to $y$, and
 * $\dfrac{\mathrm dy}{\mathrm dx}$ is the derivative of $y$ with respect to $x$.

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