Application of the Chain Rule
If $f$ has an inverse $f^{-1}$, the defining property of an inverse is that composing the two together, in either order, returns the original input unchanged:
$$f^{-1}(f(x)) = x \quad \text{for all } x \text{ in the domain of } f$$
$$f(f^{-1}(x)) = x \quad \text{for all } x \text{ in the domain of } f^{-1}$$
In other words,
$$f^{-1}\circ f = I, \qquad f \circ f^{-1} = I$$
where $I$ is the identity function, $I(x) = x$. The two compositions act as the identity on different domains in general — $f^{-1}\circ f$ is the identity on the domain of $f$, while $f\circ f^{-1}$ is the identity on the domain of $f^{-1}$, which is the range of $f$.
Differentiation of the identity $f^{-1}(f(x)) = x$
This identity itself can now be differentiated. Starting from
$$f^{-1}(f(x)) = x$$
the left-hand side is a composition, so the chain rule applies directly to it. Differentiating both sides with respect to $x$,
$$\frac{d}{dx}\Big[f^{-1}(f(x))\Big] = \frac{d}{dx}(x)$$
The right-hand side is simply $1$. The left-hand side, by the chain rule, is the derivative of $f^{-1}$ evaluated at $f(x)$, multiplied by the derivative of $f(x)$ itself:
$$(f^{-1})'(f(x)) \cdot f'(x) = 1$$
Solving for $(f^{-1})'(f(x))$,
$$(f^{-1})'(f(x)) = \frac{1}{f'(x)}$$
Writing $y = f(x)$, so that $x = f^{-1}(y)$, this becomes the familiar reciprocal form
$$\frac{dx}{dy} = \frac{1}{\dfrac{dy}{dx}}$$
The reciprocal relationship between a function’s derivative and its inverse’s derivative is therefore not a separate rule to remember — it follows directly from differentiating the composition identity using the chain rule.
Example 1: Price and demand, $q = f(p) = 100 – 4p$
Here $f'(p) = -4$. By the result just derived,
$$(f^{-1})'(q) = \frac{1}{f'(p)} = \frac{1}{-4} = -\frac{1}{4}$$
Checking directly: solving for price in terms of quantity gives $p = f^{-1}(q) = 25 – \dfrac{q}{4}$, and differentiating this directly,
$$\frac{dp}{dq} = -\frac{1}{4}$$
— which agrees exactly with the value obtained through the chain-rule derivation, without needing to invert the function explicitly at all.
Example 2: $q = f(p) = p^2$, for $p > 0$
Here $f'(p) = 2p$. By the derived result,
$$(f^{-1})'(q) = \frac{1}{f'(p)} = \frac{1}{2p}$$
Since $p = \sqrt{q}$, this can be written entirely in terms of $q$:
$$(f^{-1})'(q) = \frac{1}{2\sqrt{q}}$$
Checking directly: differentiating $p = f^{-1}(q) = \sqrt{q}$ gives $\dfrac{dp}{dq} = \dfrac{1}{2\sqrt{q}}$ — again matching exactly, confirming the chain-rule derivation without ever needing to differentiate the inverse function from scratch.