Derivative of Inverse Function Tool

Math Calculus • Derivatives

Written by STEM Calculators Team Published December 22, 2025 Updated February 24, 2026

9. Derivative Of Inverse Function Tool

Computes \(\big(f^{-1}\big)'(y_0)=\dfrac{1}{f'(x_0)}\) where \(x_0\) satisfies \(f(x_0)=y_0\), and visualizes the reflection across \(y=x\) with tangent slopes.

Inputs

Function \(f(x)\)

Supported: + − * / ^, parentheses, variable x, constants pi, e, sin cos tan, ln log(base 10), sqrt abs exp. Implicit multiplication: 2x, (x+1)(x-1), 2sin(x). Trig powers like cos^2(2x) are supported.

Target value \(y_0\)

We solve \(f(x_0)=y_0\) for \(x_0\).

Initial guess \(x_0\) (optional)

Used for Newton’s method. Leave empty for auto.

Plot center \(c\)

Graph uses \(x\) on the horizontal axis.

Graph window half-width \(w\)

Plots \(x\in[c-w,c+w]\).

Options Show grid Show inverse (reflected curve) Show tangents at corresponding points

Quick presets

Click to auto-fill and compute.

Ready

Graph

Drag to pan • wheel/pinch to zoom (wide zoom-out enabled) • shows \(f\), optional \(f^{-1}\) (reflection), and \(y=x\).

x: 0, y: 0, zoom(px/unit): 40

Result

Enter \(f(x)\), pick \(y_0\), and click Compute.

Rate this calculator

0.0 /5 (0 ratings)

Be the first to rate.

Your rating

Name (optional) Review (optional)

You can update your rating any time.

9. Derivative Of Inverse Function Tool — Theory

This tool computes the derivative of the inverse function at a point using the reciprocal-slope rule: \(\big(f^{-1}\big)'(y_0)=\dfrac{1}{f'(x_0)}\) where \(f(x_0)=y_0\).

1) When an inverse exists (locally)

A function \(f\) has an inverse (at least on an interval) if it is one-to-one there. A common sufficient condition is that \(f\) is strictly increasing or strictly decreasing on that interval.

Even if \(f\) is not globally one-to-one, it can still have a local inverse near a point \(x_0\) if \(f'(x_0)\neq 0\) and \(f\) behaves monotonically nearby (Inverse Function Theorem).

2) The inverse-derivative formula

Suppose \(y=f(x)\) and \(f\) is invertible near \(x_0\). Let \(y_0=f(x_0)\). Then:

\[ \big(f^{-1}\big)'(y_0)=\frac{1}{f'(x_0)}. \]

Derivation (short)

If \(f^{-1}\) exists near \(y_0\), then the identity holds: \[ f\big(f^{-1}(y)\big)=y. \] Differentiate both sides with respect to \(y\): \[ f'\big(f^{-1}(y)\big)\cdot (f^{-1})'(y)=1. \] Evaluate at \(y=y_0\) where \(f^{-1}(y_0)=x_0\): \[ f'(x_0)\cdot (f^{-1})'(y_0)=1 \quad\Rightarrow\quad (f^{-1})'(y_0)=\frac{1}{f'(x_0)}. \]

3) Geometric meaning: reflection and reciprocal slopes

The graph of \(f^{-1}\) is the reflection of the graph of \(f\) across the line \(y=x\). Corresponding points swap coordinates: \[ (x_0,y_0)\text{ on }f \quad\longleftrightarrow\quad (y_0,x_0)\text{ on }f^{-1}. \]

A tangent line slope of \(m=f'(x_0)\) at \((x_0,y_0)\) reflects to a tangent with slope \(1/m\) at \((y_0,x_0)\), as long as \(m\neq 0\). This is exactly the inverse-derivative formula.

4) What can go wrong?

No inverse (globally): If \(f\) is not one-to-one, there are multiple \(x\) values giving the same \(y\). A local inverse may still exist on a restricted interval.
Vertical tangent on the inverse: If \(f'(x_0)=0\), then \((f^{-1})'(y_0)=1/f'(x_0)\) blows up or is undefined.
Numerical solve issues: To compute \(x_0\), we must solve \(f(x)=y_0\). If the solver cannot find a root, you can change \(y_0\), adjust the plot window, or provide a better initial guess.

5) How the calculator computes \(x_0\)

The tool solves \(f(x_0)=y_0\) numerically using a Newton-style method:

\[ x_{n+1}=x_n-\frac{f(x_n)-y_0}{f'(x_n)}. \]

If Newton’s method stalls (for example, \(f'(x)\) is too small or the iterations do not improve), the tool attempts to bracket a root and uses bisection as a safer fallback.

6) Worked example

Let \(f(x)=x^5+x\). Find \((f^{-1})'(2)\).

Solve \(x^5+x=2\) to find \(x_0\approx 1\) (since \(1^5+1=2\)).
Compute \(f'(x)=5x^4+1\), so \(f'(1)=6\).
Therefore \((f^{-1})'(2)=\dfrac{1}{6}\).

Frequently Asked Questions

What is the derivative of an inverse function at y0?

If y0 = f(x0) and f is invertible near x0 with f'(x0) not equal to 0, then (f^{-1})'(y0) = 1 / f'(x0). The key is finding the corresponding x0 that maps to y0 under f.

How do you find x0 when you only know y0?

You solve the equation f(x) = y0 to find x0. This calculator does that numerically using an iterative method (often Newton-style) and can use a safer fallback if the iteration stalls.

Why can (f^{-1})'(y0) be undefined or extremely large?

Because (f^{-1})'(y0) = 1 / f'(x0), the value blows up or becomes undefined when f'(x0) = 0. Geometrically, a horizontal tangent on f corresponds to a vertical tangent on the inverse.

What does the reflection across y = x mean for inverse functions?

The graph of f^{-1} is the reflection of the graph of f across the line y = x. Corresponding points swap coordinates (x0, y0) on f becomes (y0, x0) on f^{-1}, and tangent slopes become reciprocal.

Rate this calculator

Frequently Asked Questions

Related calculators