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

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

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

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

Composite Function Example

Let's break it down with an example:

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

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

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

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

Therefore:

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

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:

   $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$

   • $\frac{dy}{dx}$ is the overall derivative.
   • $u$ is the inner function.
   • $\frac{dy}{du}$ is the derivative of the outer function with respect to the inner function.
   • $\frac{du}{dx}$ is the derivative of the inner function with respect to $x$.

2. Function Notation:

   $$\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$$

   • Take the derivative from the outside in ⬅️➡️.
   • If there's another function inside $g(x)$, 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 $x$.
4. Multiply: Multiply the two derivatives together.

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 $y = (x^2 + 3x - 1)^2$

1. Inner/Outer: Inner: $u = x^2 + 3x - 1$, Outer: $y = u^2$
2. Outer Derivative: $\frac{dy}{du} = 2u = 2(x^2 + 3x - 1)$
3. Inner Derivative: $\frac{du}{dx} = 2x + 3$
4. Multiply: $\frac{dy}{dx} = 2(x^2 + 3x - 1) \cdot (2x + 3)$

The Chain Rule: Example 2

Let $f(x) = e^x$ and $g(x) = \ln(x)$. Find the derivative of $y = f(g(x))$.

1. Inner/Outer: Outer: $f(x)$, Inner: $g(x)$
2. Outer Derivative: $f'(g(x)) = e^{g(x)} = e^{\ln(x)}$
3. Inner Derivative: $g'(x) = \frac{1}{x}$
4. Multiply: $f'(g(x)) \cdot g'(x) = e^{\ln(x)} \cdot \frac{1}{x}$

The Chain Rule: Example 3

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

1. Inner/Outer: Inner: $u = 5x^3 + 2x^2 + 6$, Outer: $y = 4u^2$
2. Outer Derivative: $\frac{dy}{du} = 8u = 8(5x^3 + 2x^2 + 6)$
3. Inner Derivative: $\frac{du}{dx} = 15x^2 + 4x$
4. Multiply: $\frac{dy}{dx} = 8(5x^3 + 2x^2 + 6) \cdot (15x^2 + 4x)$

The Chain Rule: Example 4

Differentiate $y = \sqrt{7x^2}$.

1. Inner/Outer: Inner: $u = 7x^2$, Outer: $y = \sqrt{u}$
2. Outer Derivative: $\frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2}(7x^2)^{-\frac{1}{2}}$
3. Inner Derivative: $\frac{du}{dx} = 14x$
4. Multiply: $\frac{dy}{dx} = \frac{1}{2}(7x^2)^{-\frac{1}{2}} \cdot (14x)$

The Chain Rule: Example 5

Try this one on your own!

$$f(x) = \cos^2(3x)$$

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

Answer:

$$f'(x) = 2(\cos 3x) \cdot -3\sin(3x)$$

📕 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

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

<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>

<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>