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Derivative of Inverse Functions

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About This MicroSim

This visualization demonstrates one of the beautiful relationships in calculus: the derivative of an inverse function is the reciprocal of the original function's derivative.

If you have a function \(f\) and its inverse \(f^{-1}\), then:

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

where \(b = f(a)\), meaning \((a, b)\) is on the graph of \(f\) and \((b, a)\) is on the graph of \(f^{-1}\).

The Geometric Intuition

Why does this work? Look at what happens when you reflect a tangent line across \(y = x\):

  • A tangent line with slope \(m\) becomes a tangent line with slope \(1/m\)
  • The reflection swaps the "rise" and "run" of the slope triangle
  • This is exactly what happens when coordinates \((a, b)\) become \((b, a)\)

Delta Moment

"Wait, if I'm rolling along \(f(x)\) at a steep angle, my reflection on \(f^{-1}\) is tilted gently? The steeper I go, the flatter my mirror-self becomes. That's... actually kind of beautiful!"

Functions Available

Function Inverse Restriction
\(f(x) = x^2\) \(f^{-1}(x) = \sqrt{x}\) \(x \geq 0\)
\(f(x) = x^3\) \(f^{-1}(x) = \sqrt[3]{x}\) none
\(f(x) = e^x\) \(f^{-1}(x) = \ln(x)\) none
\(f(x) = \sin(x)\) \(f^{-1}(x) = \arcsin(x)\) \(-\pi/2 \leq x \leq \pi/2\)

How to Use

  1. Select a function using the buttons at the bottom
  2. Drag the blue point along the curve to see how slopes change
  3. Toggle the y = x line to see or hide the reflection axis
  4. Click Animate to watch the point sweep along the curve automatically
  5. Watch the info panel to verify that the product of slopes always equals 1

What to Observe

  • The blue tangent line on \(f(x)\) and the orange tangent line on \(f^{-1}(x)\)
  • The slopes displayed in the info panel are always reciprocals
  • The product of slopes is always 1 (within rounding)
  • When \(f\) is steep, \(f^{-1}\) is shallow, and vice versa

The Mathematics

Derivation Using the Chain Rule

If \(f\) and \(f^{-1}\) are inverses, then:

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

Differentiating both sides using the chain rule:

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

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

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

If we let \(a = f^{-1}(b)\) (so \(b = f(a)\)), this becomes:

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

Example: Square Root Function

For \(f(x) = x^2\) (with \(x \geq 0\)), we have \(f^{-1}(x) = \sqrt{x}\).

  • \(f'(x) = 2x\)
  • \((f^{-1})'(x) = \frac{1}{2\sqrt{x}}\)

At the point \((2, 4)\) on \(f\): - \(f'(2) = 4\) - The corresponding point on \(f^{-1}\) is \((4, 2)\) - \((f^{-1})'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4}\)

Product: \(4 \times \frac{1}{4} = 1\)

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 geometric relationship between tangent lines on \(f\) and \(f^{-1}\)
  3. Illustrate the reflection property across \(y = x\)
  4. Apply the inverse function derivative formula to calculate specific values

Prerequisite Knowledge

  • Understanding of derivatives and tangent lines
  • Familiarity with inverse functions and the line \(y = x\)
  • Basic knowledge of exponential, logarithmic, and trigonometric functions

Suggested Activities

  1. Predict First: Before selecting a function, ask students to predict what will happen to the slope of the inverse's tangent line if the original slope is 3. Then verify with the simulation.

  2. Pattern Recognition: For each function, record several pairs of slopes. Verify that their product is always 1.

  3. Edge Cases: What happens when \(f'(a) = 0\)? When is \((f^{-1})'(b)\) undefined? Explore these cases with the simulation.

  4. Connection to Horizontal/Vertical Tangents: Find points where the tangent line on \(f\) is horizontal. What does this imply about the tangent line on \(f^{-1}\)?

Discussion Questions

  1. Why must we restrict \(f(x) = x^2\) to \(x \geq 0\) for it to have an inverse?
  2. If a function has a horizontal tangent line at some point, what can you say about its inverse at the corresponding point?
  3. Why does reflecting across \(y = x\) swap the slope to its reciprocal?
  4. How does this relationship help us find derivatives of inverse trig functions?

Assessment Questions

  1. If \(f'(3) = 5\) and \(f(3) = 7\), what is \((f^{-1})'(7)\)?
  2. Find the slope of the tangent line to \(y = \sqrt{x}\) at \(x = 9\) using the inverse function derivative formula.
  3. Explain geometrically why the tangent line to \(y = \ln(x)\) at \(x = e\) has slope \(1/e\).
  4. If the tangent line to \(f\) at \((2, 8)\) is horizontal, what can you conclude about \(f^{-1}\) at \(x = 8\)?

Embedding

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<iframe src="https://dmccreary.github.io/calculus/sims/inverse-derivative/main.html"
        height="552px" width="100%" scrolling="no"></iframe>

References