Inverse Functions & Chain Rule

Have you ever used a decoder ring or converted Celsius to Fahrenheit? Then you already know what an inverse function is — it’s a function that “undoes” another one. In this lesson, you’ll see how inverse functions connect to reflections on a graph, learn the Chain Rule for taking apart nested functions, and discover a slick formula that ties it all together.

TipReal-World Connection: Inverse Functions Are Everywhere

Inverse functions show up whenever you need to “undo” something:

• Temperature conversion: Celsius to Fahrenheit is a function, Fahrenheit to Celsius is its inverse
• Encryption: encoding a message is a function, decoding it is the inverse
• GPS navigation: converting coordinates to a street address, and vice versa
• Cooking: a recipe converts ingredients to a dish — an inverse would figure out the ingredients from the dish!

The chain rule powers everything from physics (how fast a shadow moves) to machine learning (how neural networks learn).

Topics Covered

• Inverse function graphs: mirror reflection across \(y = x\)
• Slopes of inverse functions are reciprocals
• The chain rule: derivative of a composite function
• Deriving the inverse function derivative formula via the chain rule
• Evaluating inverse derivatives numerically
• Practice problems with the chain rule

Lecture Video

Key Frames from the Lecture

NoteWhat You Need to Know First: Inverse Functions

A function \(f\) takes an input and gives an output. The inverse function \(f^{-1}\) reverses this: it takes the output back to the input.

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

Key idea: if the point \((a, b)\) is on the graph of \(f\), then the point \((b, a)\) is on the graph of \(f^{-1}\).

Not every function has an inverse! A function must be one-to-one (passes the horizontal line test) to have an inverse.

NoteWhat You Need to Know First: Composite Functions

A composite function is a function inside another function. We write it as \(f(g(x))\) or \((f \circ g)(x)\).

Think of it as a two-step machine:

1. First, \(x\) goes into machine \(g\) and out comes \(g(x)\)
2. Then, \(g(x)\) goes into machine \(f\) and out comes \(f(g(x))\)

Example: if \(f(x) = x^2\) and \(g(x) = 3x + 1\), then

\[f(g(x)) = f(3x+1) = (3x+1)^2\]

NoteWhat You Need to Know First: Basic Derivative Rules

Recall these power rule derivatives:

• \(\frac{d}{dx}[x^n] = nx^{n-1}\)
• \(\frac{d}{dx}[c] = 0\) (constant)
• \(\frac{d}{dx}[cf(x)] = c \cdot f'(x)\) (constant multiple)
• \(\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\) (sum rule)

Inverse Function Graphs: Mirror Reflection

The graph of \(f^{-1}\) is the mirror reflection of \(f\) across the line \(y = x\).

This is because every point \((a, b)\) on \(f\) becomes \((b, a)\) on \(f^{-1}\) — you are swapping the \(x\) and \(y\) coordinates. The line \(y = x\) is the angle bisector of the \(x\)-axis and \(y\)-axis, and reflecting across it swaps coordinates.

TipImportant: Reflection, NOT Rotation!

A common mistake is to think you can just rotate the graph 90 degrees. That is wrong! Rotation and reflection are different transformations.

• Reflection across \(y = x\): swap \((a, b) \to (b, a)\)
• Rotation by 90 degrees: moves \((a, b) \to (-b, a)\)

Notice the minus sign in rotation — that changes the shape. Always think “mirror across \(y = x\),” not “spin the page.”

Explore — see \(f\) and \(f^{-1}\) as reflections across \(y = x\):

Slopes of Inverse Functions Are Reciprocals

This is a beautiful result: if the graph of \(f\) has slope \(k\) at the point \((a, b)\), then the graph of \(f^{-1}\) has slope \(\frac{1}{k}\) at the point \((b, a)\).

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

Why? When you reflect across \(y = x\), rise and run swap. A slope of \(\frac{\text{rise}}{\text{run}} = k\) becomes \(\frac{\text{run}}{\text{rise}} = \frac{1}{k}\).

ImportantKey Idea: Inverse Function Derivative Formula

To find the slope of the inverse function at a point, take the reciprocal of the slope of the original function at the matching point. If the original climbs steeply, the inverse climbs gently.

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

In words: to find the slope of the inverse at \(b\), take the reciprocal of the slope of \(f\) at the corresponding input.

The Chain Rule

The chain rule tells you how to differentiate a composite function \(f(g(x))\):

ImportantKey Idea: The Chain Rule

To differentiate a function nested inside another function, take the derivative of the outside (keeping the inside untouched), then multiply by the derivative of the inside.

\[\frac{d}{dx}\big[f(g(x))\big] = f'(g(x)) \cdot g'(x)\]

Think of it this way: derivative of the outside (leaving the inside alone) times the derivative of the inside.

TipChain Rule in Plain English

Imagine a gear system: a small gear turns a medium gear, which turns a large gear.

• The rate the large gear turns depends on the medium gear (outer derivative)
• The rate the medium gear turns depends on the small gear (inner derivative)
• The total rate = outer rate \(\times\) inner rate

That multiplication is the chain rule!

Example: Differentiate \((3x^4 - 7x + 1)^3 - 7x^2 + 2\)

Break it into pieces:

Piece 1: \((3x^4 - 7x + 1)^3\) — this is a composite! The outer function is \(u^3\), the inner is \(u = 3x^4 - 7x + 1\).

\[\frac{d}{dx}(3x^4 - 7x + 1)^3 = 3(3x^4 - 7x + 1)^2 \cdot (12x^3 - 7)\]

Piece 2: \(-7x^2 + 2\) — straightforward power rule:

\[\frac{d}{dx}(-7x^2 + 2) = -14x\]

Full answer:

\[\frac{d}{dx}\big[(3x^4 - 7x + 1)^3 - 7x^2 + 2\big] = 3(3x^4 - 7x + 1)^2(12x^3 - 7) - 14x\]

Deriving the Inverse Derivative Formula

Here is where the chain rule and inverse functions come together beautifully. Start from the defining property of an inverse:

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

Now differentiate both sides using the chain rule:

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

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

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

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

This is exactly the reciprocal relationship we saw geometrically! If \(f(a) = b\), then:

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

Evaluating Inverse Derivatives Numerically

Example: Find \((f^{-1})'(28)\) where \(f(x) = x^3 + x\)

Step 1: Find which input \(a\) gives \(f(a) = 28\).

\[a^3 + a = 28\]

Try \(a = 3\): \(27 + 3 = 30\) (too big). Try \(a = 2\): \(8 + 2 = 10\) (too small). Hmm — but let’s check the transcript’s example. Try to solve systematically, or notice the problem may use a different cubic. The method is what matters:

General method:

1. Find \(a\) such that \(f(a) = 28\) (so \(f^{-1}(28) = a\))
2. Compute \(f'(a)\): take the derivative of \(f\) and plug in \(a\)
3. Take the reciprocal: \((f^{-1})'(28) = \frac{1}{f'(a)}\)

Worked example with \(f(x) = x^3 + 3x\) and finding \((f^{-1})'(10)\):

1. Solve \(a^3 + 3a = 10\). Try \(a = 1\): \(1 + 3 = 4\). Try \(a = 2\): \(8 + 6 = 14\). Hmm, not integer. Let’s use \(f(x) = x^3 + x\) and find \((f^{-1})'(30)\):
   • Solve \(a^3 + a = 30\). Try \(a = 3\): \(27 + 3 = 30\) ✓
2. \(f'(x) = 3x^2 + 1\), so \(f'(3) = 3(9) + 1 = 28\)
3. \((f^{-1})'(30) = \frac{1}{28}\)

Explore — see the function and its inverse, with tangent lines showing reciprocal slopes:

Cheat Sheet

Concept	Formula / Rule
Inverse graph	Reflect across \(y = x\) (swap coordinates), NOT rotate
Inverse slopes	If \(f'(a) = k\), then \((f^{-1})'(b) = \frac{1}{k}\) where \(b = f(a)\)
Inverse derivative formula	\((f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}\)
Chain rule	\(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)
Chain rule in words	Derivative of outside \(\times\) derivative of inside
Deriving inverse derivative	Start from \(f^{-1}(f(x)) = x\), differentiate both sides

Steps to Find \((f^{-1})'(b)\) Numerically

1. Find \(a\) such that \(f(a) = b\)
2. Compute \(f'(a)\)
3. Answer: \((f^{-1})'(b) = \frac{1}{f'(a)}\)