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

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

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:

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

Thus $$\frac {f^{-1} \left({y + h}\right) - f^{-1} \left({y}\right)} {h} \to \frac 1 {D f \left({x}\right)}$$ as $$h \to 0$$.


 * Suppose $$f$$ is strictly decreasing.

Exactly the same argument applies.

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

$$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$ where $$\frac{dx}{dy}$$ is the derivative of $$x$$ with respect to $$y$$ and $$\frac{dy}{dx}$$ 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.