Deriving the Inverse Function Derivative Formula

Why the Inverse Derivative Formula Has a Reciprocal

A function and its inverse undo each other: the inverse takes an output of the original function back to its input. Their rates of change are therefore closely related. If the original function changes rapidly near an input, its inverse changes more slowly near the corresponding output; if the original function changes slowly, the inverse must change more rapidly.

The exact relationship depends on matching the two corresponding inputs correctly. The formula is not simply the reciprocal of the derivative at the same number. We will derive the relationship from composition and then use a difference quotient to see the conditions behind it.

The Formula and Its Conditions

Let \(f\) be one-to-one on an interval, and write \(g=f^{-1}\) for its inverse on the corresponding range. If \(x_0\) is an input to \(f\), set \(a=f(x_0)\). Thus \(g(a)=x_0\).

Strict monotonicity makes \(f\) one-to-one, so the inverse exists on its range. Continuity and strict monotonicity ensure that the inverse is continuous at \(a\). The nonzero derivative is essential: it allows us to take the reciprocal of the original slope and ensures the inverse has a finite derivative there.

The formula is often written in terms of the inverse input \(a\):

The point \(f^{-1}(a)\) is the input to the original function that produces \(a\). The derivative in the denominator must be evaluated at that point.

Deriving the Formula from Composition

Because \(f\) and \(g=f^{-1}\) undo each other, their composition is the identity function. In particular, \(f(g(y))=y\) for \(y\) in the range of \(f\). If both functions are differentiable at the relevant points, the chain rule gives

Setting \(y=a\) and solving for \(g'(a)\) gives the desired relationship, provided \(f'(g(a))\ne 0\). This chain-rule calculation explains the reciprocal: the derivative of the identity function is \(1\), so the two rates of change must multiply to \(1\).

A Difference-Quotient Justification

The quotient argument both proves the theorem under its stated conditions and makes the input matching explicit. Write \(g=f^{-1}\), so \(g(a)=x_0\). For \(y\ne a\), the definition of the derivative gives

Since \(f(g(y))=y\) and \(f(g(a))=a\), the denominator can be rewritten using the original function:

As \(y\) approaches \(a\), continuity of \(g\) gives \(g(y)\to g(a)=x_0\). Since \(f\) is one-to-one, \(y\ne a\) implies \(g(y)\ne g(a)\). We can therefore take the reciprocal of the difference quotient for \(f\):

The denominator on the right approaches \(f'(x_0)\), which is nonzero. Taking the limit therefore gives

This completes the justification: the inverse derivative is the reciprocal because its difference quotient is the reciprocal of the original function’s difference quotient at the matching input.

Worked Examples

When the Formula Does Not Apply

If \(f'(x_0)=0\), the displayed reciprocal is undefined. This is not just a technical issue: the inverse may fail to have a finite derivative at the corresponding output. For example, \(f(x)=x^3\) is strictly increasing and has inverse \(g(y)=\sqrt[3]{y}\). At \(x_0=0\), \(f'(0)=0\), and the inverse has no finite derivative at \(y=0\): for \(y\ne 0\),

which grows without bound as \(y\to 0\). Thus the usual finite inverse derivative formula cannot be used there.

Why the Conditions Matter

The inverse derivative relationship is useful whenever a function is easier to differentiate than its inverse. It also gives a quick interpretation of slopes: at matching points, the original and inverse slopes multiply to \(1\). If the original slope is negative, the inverse slope is negative; if the original slope is large in magnitude, the inverse slope is small in magnitude.

Before applying the formula, check three things:

• The original function has an inverse on the interval being used.
• You have identified \(x_0=f^{-1}(a)\), so that \(f(x_0)=a\).
• The derivative \(f'(x_0)\) exists and is nonzero.

These checks keep the formula tied to its actual meaning: it relates the inverse’s rate of change at an output to the original function’s rate of change at the input that produced that output.