Derivative of Inverse Function

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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:

$\forall x \in I^o: D f \left({x}\right) > 0$


$\forall x \in I^o: D f \left({x}\right) < 0$


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


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

strictly increasing on $I$ (if $\forall x \in I^o: D f \left({x}\right) > 0$)


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

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


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


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

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:

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


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

as $h \to 0$.

Suppose $f$ is strictly decreasing.

Exactly the same argument applies.


Style Note

Leibniz's notation for derivatives $\left (\dfrac{\mathrm d y}{\mathrm d x}\right)$ allows for a particularly elegant statement of this rule:

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


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