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Graphical Inverse Derivatives

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About This MicroSim

This visualization demonstrates one of calculus's most elegant relationships: the derivative of an inverse function equals the reciprocal of the original function's derivative. In mathematical notation:

\[\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}\]

When you drag a point along f(x) on the left graph, the corresponding point on \(f^{-1}(x)\) updates on the right graph. The tangent lines at these corresponding points always have slopes that are reciprocals of each other, meaning their product equals 1.

Delta Moment

"Check this out: if I'm climbing a slope of 2 on f(x), the same spot on the inverse has me at slope 1/2. Flip it and reverse it - literally! The slopes are reciprocals, and 2 times 1/2 is... wait for it... 1!"

The Key Insight

If the point \((a, b)\) is on \(f(x)\), then the point \((b, a)\) is on \(f^{-1}(x)\) - the coordinates swap. The tangent line slopes at these corresponding points multiply to exactly 1:

\[m_f \cdot m_{f^{-1}} = 1\]

Why Does This Work?

Think about it geometrically: reflecting a line across \(y = x\) inverts its slope. If a tangent has slope \(m\), its reflection has slope \(1/m\). Since \(f^{-1}(x)\) is the reflection of \(f(x)\) across \(y = x\), their tangent lines at corresponding points must have reciprocal slopes.

How to Use

  1. Select a Function: Use the dropdown to choose different function families (Cubic, Square Root, Exponential, Linear)
  2. Drag the Point: Click and drag on the left graph to move the point along f(x)
  3. Use the Slider: Adjust the x-value precisely using the slider
  4. Toggle y = x Line: Show or hide the dashed reference line
  5. Toggle Slopes: Show or hide the slope calculation panel
  6. Animate: Watch the point trace along the curve automatically

What to Observe

  • The blue tangent line on f(x) and the orange tangent line on \(f^{-1}(x)\) have reciprocal slopes
  • The slope panel shows \(m_f \times m_{inv} = 1\) at every point
  • The dashed \(y = x\) line shows the mirror relationship between the curves
  • As slope increases on f(x), it decreases on \(f^{-1}(x)\) and vice versa

Lesson Plan

Learning Objectives

After using this MicroSim, students will be able to:

  1. Explain why the derivative of an inverse function equals the reciprocal of the original derivative (Bloom Level 2: Understand)
  2. Interpret the graphical relationship between tangent lines on f(x) and \(f^{-1}(x)\)
  3. Demonstrate the reciprocal slope property by identifying corresponding points and their tangent slopes

Prerequisite Knowledge

  • Understanding of derivatives as slopes of tangent lines
  • Knowledge of inverse functions and their graphical relationship (reflection across y = x)
  • Basic familiarity with exponential, logarithmic, and power functions

Suggested Activities

  1. Prediction Challenge: Before moving the point, students predict: "If the slope on f(x) is 3, what will the slope be on \(f^{-1}(x)\)?" Verify with the simulation.

  2. Slope Detective: For each function type, find the point where both slopes equal 1. What special property does this point have? (It lies on the line y = x!)

  3. Edge Cases: Explore what happens as slopes approach 0 or infinity. For the square root function, observe how a very small slope on one curve corresponds to a very large slope on the other.

  4. Formula Connection: After building intuition graphically, derive the formula algebraically using implicit differentiation on \(f(f^{-1}(x)) = x\).

Discussion Questions

  1. Why must the product of the slopes always equal 1?
  2. What happens at a point where f'(x) = 0? Can the inverse have a tangent line there?
  3. How does the reflection relationship across y = x explain the reciprocal slopes?
  4. For f(x) = \(e^x\), the derivative at any point equals the y-coordinate. How does this connect to the inverse?

Assessment Questions

  1. If f(x) = \(x^3\) and you're at the point (2, 8), what is the slope of the tangent to \(f^{-1}(x)\) at (8, 2)?
  2. For f(x) = \(\sqrt{x}\) at x = 4, the slope is 1/4. What is the slope of \(f^{-1}(x)\) at the corresponding point?
  3. Explain in your own words why inverse function derivatives are reciprocals.

Mathematical Background

The Inverse Function Derivative Formula

Starting from \(f(f^{-1}(x)) = x\), differentiate both sides using the chain rule:

\[f'(f^{-1}(x)) \cdot (f^{-1})'(x) = 1\]

Solving for \((f^{-1})'(x)\):

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

Example: Exponential and Logarithm

For \(f(x) = e^x\) with \(f^{-1}(x) = \ln(x)\):

  • At x = 1 on \(f(x) = e^x\): the point is \((1, e)\) and \(f'(1) = e\)
  • The corresponding point on \(\ln(x)\) is \((e, 1)\)
  • The slope of \(\ln(x)\) at \(x = e\) is \(1/e\)
  • Product: \(e \cdot \frac{1}{e} = 1\)

Embedding

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<iframe src="https://dmccreary.github.io/calculus/sims/graphical-inverse-derivatives/main.html"
        height="525px" width="100%" scrolling="no"></iframe>

References