The graph of \(f^{-1}\) is the graph of \(f\) reflected across the line \(y=x\). Reflection across \(y=x\) swaps horizontal and vertical change, and at corresponding points, tangent slopes are therefore reciprocals. That geometric picture leads directly to the derivative formula.

The rule and the evaluation point

Suppose \(f\) is one-to-one and differentiable, and suppose \(f'\big(f^{-1}(a)\big)\neq0\). Then

\[\big(f^{-1}\big)'(a) = \frac{1}{f'\big(f^{-1}(a)\big)}\]

The formula follows from the identity \(f\big(f^{-1}(x)\big)=x\). Differentiate both sides to get \(f'\big(f^{-1}(x)\big) \cdot \big(f^{-1}\big)'(x) = 1\), then solve for the inverse derivative.

The main difficulty is usually not the reciprocal. It is finding where \(f'\) should be evaluated. Suppose \(f(x)=x^3+x\). Since \(f(1)=2\), we know \(f^{-1}(2)=1\), and therefore

\[\big(f^{-1}\big)'(2) = \frac{1}{f'(1)} = \frac{1}{3(1)^2+1} = \frac14\]

We never needed a formula for \(f^{-1}\).

Matching corresponding points

The dark curve is f and the pale one is its reflection across the dashed diagonal, drawn by swapping coordinates rather than by inverting anything. A point sits at (b, f(b)) and its mirror at (f(b), b); the two short lines are the tangents there. The panel gives f'(b), its reciprocal, and their product, which is 1 wherever both exist. Where an inverse derivative has a formula worth memorizing, the last row evaluates that formula independently and reports the gap. For x³ + x there is no formula to compare against, which is the case the rule was built for.

The visualization shows \(f\) and its reflection across \(y=x\). A point \((b,f(b))\) on the original graph corresponds to \((f(b),b)\) on the inverse, and the tangent slopes at those two points are reciprocals wherever both derivatives exist.

This is useful when a problem provides a table rather than a formula. If you are asked for \(\big(f^{-1}\big)'(4)\), look for the row where \(f(x)=4\). That row gives \(f^{-1}(4)\), and the derivative \(f'\) must be evaluated at that input. Looking for \(x=4\) instead is the common mistake.

Inverse trigonometric derivatives

The inverse trigonometric derivative formulas can be derived from the same rule. For \(f=\sin\) on the interval where sine is one-to-one, \((\arcsin)'(a) = \tfrac{1}{\cos(\arcsin a)}\), and using the Pythagorean identity, \(\cos(\arcsin a) = \sqrt{1-a^2}\), so \((\arcsin)'(a) = \tfrac{1}{\sqrt{1-a^2}}\).

For tangent, \((\arctan)'(a) = \tfrac{1}{\sec^2(\arctan a)}\), and since \(\sec^2\theta=1+\tan^2\theta\), we get \((\arctan)'(a) = \tfrac{1}{1+a^2}\). These formulas are consequences of the inverse-function rule and trigonometric identities.

When the reciprocal does not exist

Take \(f(x)=x^3\), whose inverse is \(f^{-1}(x)=x^{1/3}\). At the origin, \(f'(0)=0\), so the reciprocal formula would require division by zero.

The graph explains why. The tangent to \(x^3\) at the origin is horizontal, and reflecting that tangent across \(y=x\) produces a vertical line. The inverse therefore has a vertical tangent at the corresponding point, not a finite derivative.

The condition \(f'\big(f^{-1}(a)\big)\neq0\) is part of the theorem for a reason.