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The Chain Rule

Hannah Hill

Hannah Hill

6 min read

Next Topic - Implicit Differentiation

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Study Guide Overview

This study guide covers the Chain Rule for differentiating composite functions. It explains how to identify inner and outer functions, apply the chain rule using both Leibniz and function notations, and work through practice problems. The guide also emphasizes common mistakes and provides practice questions for the final exam.

#Unit 3: Differentiation - Mastering the Chain Rule 🚀

Welcome to Unit 3! This unit is all about mastering different differentiation techniques, and we're kicking it off with the Chain Rule, a fundamental concept for handling composite functions. Let's dive in!

#🔄 Composite Functions

Key Concept

Composite functions are functions inside other functions. Think of it like a set of Russian nesting dolls!

Given two functions, f(x)f(x)f(x) and g(x)g(x)g(x), the composite function (f∘g)(x)(f \circ g)(x)(f∘g)(x) is formed by applying fff to the output of ggg. Mathematically:

(f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))(f∘g)(x)=f(g(x))

  • g(x)g(x)g(x) is the inner function.
  • f(x)f(x)f(x) is the outer function.

#Composite Function Example

Let's break it down with an example:

f(x)=x2\textcolor{red}{f(x) = x^2}f(x)=x2

g(x)=3x+1\textcolor{blue}{g(x) = 3x + 1}g(x)=3x+1

For f(g(x))f(g(x))f(g(x)), we have:

  • The inner function, g(x)g(x)g(x), takes xxx, multiplies it by 3, and adds 1. - The outer function, f(x)f(x)f(x), takes the result from g(x)g(x)g(x) and squares it.

Therefore:

f(g(x))=(3x+1)2f(g(x)) = \textcolor{red}{(\textcolor{blue}{3x + 1})^2}f(g(x))=(3x+1)2

Understanding this is key to using the Chain Rule!


#🔗 Definition of The Chain Rule

The Chain Rule is your go-to method for differentiating composite functions. It's used a lot, so make sure you know it well!

There are two common notations:

  1. Leibniz Notation:

    dydx=dydu⋅dudx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}dxdy​=dudy​⋅dxdu​

    • dydx\frac{dy}{dx}dxdy​ is the overall derivative.
    • uuu is the inner function.
    • dydu\frac{dy}{du}dudy​ is the derivative of the outer function with respect to the inner function.
    • dudx\frac{du}{dx}dxdu​ is the derivative of the inner function with respect to xxx.
  2. Function Notation:

    ddx(f(g(x)))=f′(g(x))⋅g′(x)\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)dxd​(f(g(x)))=f′(g(x))⋅g′(x)

    • Take the derivative from the outside in ⬅️➡️.
    • If there's another function inside g(x)g(x)g(x), you’d repeat the process.

#🪜 Steps to Chain Rule

  1. Identify: Define your inner and outer functions.
  2. Outer Derivative: Take the derivative of the outer function, leaving the inner function as is.
  3. Inner Derivative: Take the derivative of the inner function with respect to xxx.
  4. Multiply: Multiply the two derivatives together.
Memory Aid

Think of the Chain Rule as peeling an onion: you differentiate the outer layer, then the next layer, and so on, multiplying each derivative as you go.

markdown-image

Image Courtesy of Geeks for Geeks


# 🔢 The Chain Rule: Practice Problems

Let's solidify your understanding with some examples.

#The Chain Rule: Example 1

Find the derivative of y=(x2+3x−1)2y = (x^2 + 3x - 1)^2y=(x2+3x−1)2

  1. Inner/Outer: Inner: u=x2+3x−1u = x^2 + 3x - 1u=x2+3x−1, Outer: y=u2y = u^2y=u2
  2. Outer Derivative: dydu=2u=2(x2+3x−1)\frac{dy}{du} = 2u = 2(x^2 + 3x - 1)dudy​=2u=2(x2+3x−1)
  3. Inner Derivative: dudx=2x+3\frac{du}{dx} = 2x + 3dxdu​=2x+3
  4. Multiply: dydx=2(x2+3x−1)⋅(2x+3)\frac{dy}{dx} = 2(x^2 + 3x - 1) \cdot (2x + 3)dxdy​=2(x2+3x−1)⋅(2x+3)

#The Chain Rule: Example 2

Let f(x)=exf(x) = e^xf(x)=ex and g(x)=ln⁡(x)g(x) = \ln(x)g(x)=ln(x). Find the derivative of y=f(g(x))y = f(g(x))y=f(g(x)).

  1. Inner/Outer: Outer: f(x)f(x)f(x), Inner: g(x)g(x)g(x)
  2. Outer Derivative: f′(g(x))=eg(x)=eln⁡(x)f'(g(x)) = e^{g(x)} = e^{\ln(x)}f′(g(x))=eg(x)=eln(x)
  3. Inner Derivative: g′(x)=1xg'(x) = \frac{1}{x}g′(x)=x1​
  4. Multiply: f′(g(x))⋅g′(x)=eln⁡(x)⋅1xf'(g(x)) \cdot g'(x) = e^{\ln(x)} \cdot \frac{1}{x}f′(g(x))⋅g′(x)=eln(x)⋅x1​

#The Chain Rule: Example 3

Find the derivative of y=4(5x3+2x2+6)2y = 4(5x^3 + 2x^2 + 6)^2y=4(5x3+2x2+6)2.

  1. Inner/Outer: Inner: u=5x3+2x2+6u = 5x^3 + 2x^2 + 6u=5x3+2x2+6, Outer: y=4u2y = 4u^2y=4u2
  2. Outer Derivative: dydu=8u=8(5x3+2x2+6)\frac{dy}{du} = 8u = 8(5x^3 + 2x^2 + 6)dudy​=8u=8(5x3+2x2+6)
  3. Inner Derivative: dudx=15x2+4x\frac{du}{dx} = 15x^2 + 4xdxdu​=15x2+4x
  4. Multiply: dydx=8(5x3+2x2+6)⋅(15x2+4x)\frac{dy}{dx} = 8(5x^3 + 2x^2 + 6) \cdot (15x^2 + 4x)dxdy​=8(5x3+2x2+6)⋅(15x2+4x)

#The Chain Rule: Example 4

Differentiate y=7x2y = \sqrt{7x^2}y=7x2​.

  1. Inner/Outer: Inner: u=7x2u = 7x^2u=7x2, Outer: y=uy = \sqrt{u}y=u​
  2. Outer Derivative: dydu=12u−12=12(7x2)−12\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2}(7x^2)^{-\frac{1}{2}}dudy​=21​u−21​=21​(7x2)−21​
  3. Inner Derivative: dudx=14x\frac{du}{dx} = 14xdxdu​=14x
  4. Multiply: dydx=12(7x2)−12⋅(14x)\frac{dy}{dx} = \frac{1}{2}(7x^2)^{-\frac{1}{2}} \cdot (14x)dxdy​=21​(7x2)−21​⋅(14x)

#The Chain Rule: Example 5

Try this one on your own!

f(x)=cos⁡2(3x)f(x) = \cos^2(3x)f(x)=cos2(3x)

Hint: Inner function is u=cos⁡(3x)u = \cos(3x)u=cos(3x), outer function is f(x)=u2f(x) = u^2f(x)=u2.

Answer:

f′(x)=2(cos⁡3x)⋅−3sin⁡(3x)f'(x) = 2(\cos 3x) \cdot -3\sin(3x)f′(x)=2(cos3x)⋅−3sin(3x)

Common Mistake

Remember to apply the chain rule to every layer of the function. Don't forget the derivative of the innermost function!


#📕 Closing

You've made it through the first key topic of Unit 3! The Chain Rule will be your trusty sidekick throughout your AP Calculus journey. Keep practicing, and you'll become a Chain Rule master! 🌟

#🎯 Final Exam Focus

  • High-Priority: Chain Rule is everywhere! Expect to see it in multiple-choice and free-response questions.
  • Connections: Often combined with other differentiation rules (Product, Quotient).
  • Common Pitfalls: Forgetting the inner derivative, not applying the chain rule to every layer.
  • Time Management: Practice identifying inner and outer functions quickly.

# 📝 Practice Questions

Practice Question

#Multiple Choice Questions

  1. If y=sin⁡(x2)y = \sin(x^2)y=sin(x2), then dydx\frac{dy}{dx}dxdy​ is: (A) cos⁡(x2)\cos(x^2)cos(x2) (B) 2xcos⁡(x2)2x\cos(x^2)2xcos(x2) (C) −cos⁡(x2)-\cos(x^2)−cos(x2) (D) −2xcos⁡(x2)-2x\cos(x^2)−2xcos(x2)

  2. The derivative of f(x)=exf(x) = e^{\sqrt{x}}f(x)=ex​ is: (A) ex2x\frac{e^{\sqrt{x}}}{2\sqrt{x}}2x​ex​​ (B) exe^{\sqrt{x}}ex​ (C) exx\frac{e^{\sqrt{x}}}{\sqrt{x}}x​ex​​ (D) 2xex2\sqrt{x}e^{\sqrt{x}}2x​ex​

#Free Response Question

Question:

Consider the function h(x)=4+f(x)h(x) = \sqrt{4 + f(x)}h(x)=4+f(x)​, where f(1)=5f(1) = 5f(1)=5 and f′(1)=−3f'(1) = -3f′(1)=−3.

(a) Find h(1)h(1)h(1).

(b) Find h′(x)h'(x)h′(x).

(c) Find h′(1)h'(1)h′(1).

Scoring Breakdown:

(a) 1 point: Correct evaluation of h(1)h(1)h(1).

<math-inline>h(1) = \sqrt{4 + f(1)} = \sqrt{4 + 5} = \sqrt{9} = 3</math-inline>

(b) 2 points: 1 point for applying the chain rule correctly, 1 point for the derivative of the square root.

<math-inline>h'(x) = \frac{1}{2}(4 + f(x))^{-\frac{1}{2}} \cdot f'(x) = \frac{f'(x)}{2\sqrt{4 + f(x)}}</math-inline>

(c) 2 points: 1 point for substituting the values correctly, 1 point for the correct answer.

<math-inline>h'(1) = \frac{f'(1)}{2\sqrt{4 + f(1)}} = \frac{-3}{2\sqrt{4 + 5}} = \frac{-3}{2\sqrt{9}} = \frac{-3}{2 \cdot 3} = -\frac{1}{2}</math-inline>

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Question 1 of 9

🚀 Given f(x)=x3f(x) = x^3f(x)=x3 and g(x)=2x+1g(x) = 2x + 1g(x)=2x+1, what is the inner function in the composite function f(g(x))f(g(x))f(g(x))?

x3x^3x3

2x+12x+12x+1

(2x+1)3(2x+1)^3(2x+1)3

3x23x^23x2