Or in Leibniz notation:

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

where $u = g(x)$.

💡 Memory Trick: "Derivative of the outside times derivative of the inside"

Understanding Composite Functions

A composite function is a function within another function.

Examples of Composite Functions

1. $f(x) = (x^2 + 1)^5$ — The function $x^2 + 1$ is inside the power of 5
2. $f(x) = \sin(3x)$ — The function $3x$ is inside the sine function
3. $f(x) = e^{x^2}$ — The function $x^2$ is inside the exponential
4. $f(x) = \sqrt{x^3 - 2x}$ — The function is inside the square root

The "Outside-Inside" Method

Step-by-Step Process

1. Identify the outside function and the inside function
2. Differentiate the outside function (leave the inside alone)
3. Multiply by the derivative of the inside function

Example Walkthrough

Find the derivative of $f(x) = (x^2 + 3x)^4$

Step 1: Identify

• Outside function: $u^4$ (something to the 4th power)
• Inside function: $u = x^2 + 3x$

Step 2: Differentiate the outside

• Derivative of $u^4$ is $4u^3$
• Replace $u$ with $(x^2 + 3x)$: $4(x^2 + 3x)^3$

Step 3: Multiply by derivative of inside

• Derivative of $(x^2 + 3x)$ is $2x + 3$
• Final answer: $f'(x) = 4(x^2 + 3x)^3(2x + 3)$

Common Chain Rule Patterns

Pattern 1: Powers of Functions

If $f(x) = [g(x)]^n$, then $f'(x) = n[g(x)]^{n-1} \cdot g'(x)$

Example: $(3x^2 - 1)^7$

• Answer: $7(3x^2 - 1)^6 \cdot 6x = 42x(3x^2 - 1)^6$

Pattern 2: Square Roots

If $f(x) = \sqrt{g(x)}$, then $f'(x) = \frac{g'(x)}{2\sqrt{g(x)}}$

Example: $\sqrt{x^2 + 1}$

• Answer: $\frac{2x}{2\sqrt{x^2 + 1}} = \frac{x}{\sqrt{x^2 + 1}}$

Pattern 3: Functions with Coefficients

If $f(x) = g(ax + b)$, then $f'(x) = a \cdot g'(ax + b)$

Example: $\sin(5x)$

• Answer: $5\cos(5x)$

⚠️ Common Mistakes

Mistake 1: Forgetting the Chain Rule

❌ Wrong: $\frac{d}{dx}[(2x + 1)^3] = 3(2x + 1)^2$ ✅ Right: $\frac{d}{dx}[(2x + 1)^3] = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2$

Mistake 2: Not Simplifying

While $(3x^2 - 1)^6 \cdot 6x$ is correct, it's better to write $6x(3x^2 - 1)^6$ (constants first).

Mistake 3: Confusing Inside and Outside

Make sure you identify which function is inside and which is outside!

🎯 When to Use the Chain Rule

Use the Chain Rule whenever you see:

• A function raised to a power (other than just $x$)
• Trig functions with anything other than just $x$ inside
• Exponentials with anything other than just $x$ in the exponent
• Square roots or other roots of expressions
• Any function within another function

💡 Quick Test: If you can't apply the Power Rule, Product Rule, or Quotient Rule directly, you probably need the Chain Rule!

Multiple Compositions

Sometimes you need to apply the Chain Rule more than once!

Example: $f(x) = \sin^2(3x) = [\sin(3x)]^2$

This has TWO layers:

1. Outside: squaring function
2. Middle: sine function
3. Inside: $3x$

Solution:

• Derivative of square: $2\sin(3x)$
• Times derivative of sine: $2\sin(3x) \cdot \cos(3x)$
• Times derivative of $3x$: $2\sin(3x)\cos(3x) \cdot 3$
• Answer: $f'(x) = 6\sin(3x)\cos(3x)$

📝 Practice Strategy

1. Circle the inside function
2. Box the outside function
3. Write "outside' × inside'" as a reminder
4. Apply the formula step by step
5. Simplify your answer

Let $u = 3x^2 - 5$, then $y = u^7$

$\frac{dy}{du} = 7u^6$ $\frac{du}{dx} = 6x$

$f'(x) = 7(3x^2 - 5)^6 \cdot 6x = 42x(3x^2 - 5)^6$

Part (b): Let $u = 4x$, then $y = \sin u$

$\frac{dy}{du} = \cos u$ $\frac{du}{dx} = 4$

$g'(x) = \cos(4x) \cdot 4 = 4\cos(4x)$

Part (c): Let $u = x^2 + 1$, then $y = e^u$

$\frac{dy}{du} = e^u$ $\frac{du}{dx} = 2x$

$h'(x) = e^{x^2+1} \cdot 2x = 2xe^{x^2+1}$

Step 2: Identify the functions

Outside function: $u^{1/2}$

Inside function: $u = 4x^2 + 9$

Step 3: Apply the Chain Rule

$\frac{dy}{dx} = \frac{1}{2}(4x^2 + 9)^{-1/2} \cdot \frac{d}{dx}[4x^2 + 9]$

Step 4: Find the derivative of the inside

$\frac{d}{dx}[4x^2 + 9] = 8x$

Step 5: Combine and simplify

$\frac{dy}{dx} = \frac{1}{2}(4x^2 + 9)^{-1/2} \cdot 8x$

$= 4x(4x^2 + 9)^{-1/2}$

$= \frac{4x}{\sqrt{4x^2 + 9}}$

Answer: $\displaystyle\frac{dy}{dx} = \frac{4x}{\sqrt{4x^2 + 9}}$

$\frac{dy}{du} = \frac{1}{2}u^{-1/2} = \frac{1}{2\sqrt{u}}$

Now find $\frac{du}{dx}$ where $u = \cos(3x^2)$:

Let $v = 3x^2$, then $u = \cos v$

$\frac{du}{dv} = -\sin v$ $\frac{dv}{dx} = 6x$

$\frac{du}{dx} = -\sin(3x^2) \cdot 6x = -6x\sin(3x^2)$

Combine:

$\frac{dy}{dx} = \frac{1}{2\sqrt{\cos(3x^2)}} \cdot (-6x\sin(3x^2))$

$= \frac{-6x\sin(3x^2)}{2\sqrt{\cos(3x^2)}}$

$= \frac{-3x\sin(3x^2)}{\sqrt{\cos(3x^2)}}$

So $g(x) = u \cdot v$

Step 2: Apply the Product Rule

$g'(x) = u'v + uv'$

Step 3: Find $u'$ using the Chain Rule

$u = (x^2 + 1)^3$

$u' = 3(x^2 + 1)^2 \cdot 2x = 6x(x^2 + 1)^2$

Step 4: Find $v'$ using the Chain Rule

$v = (2x - 5)^4$

$v' = 4(2x - 5)^3 \cdot 2 = 8(2x - 5)^3$

Step 5: Substitute into Product Rule

$g'(x) = [6x(x^2 + 1)^2][(2x - 5)^4] + [(x^2 + 1)^3][8(2x - 5)^3]$

Step 6: Factor out common terms

Factor out $(x^2 + 1)^2(2x - 5)^3$:

$g'(x) = (x^2 + 1)^2(2x - 5)^3[6x(2x - 5) + 8(x^2 + 1)]$

Simplify inside the brackets:

$= (x^2 + 1)^2(2x - 5)^3[12x^2 - 30x + 8x^2 + 8]$

$= (x^2 + 1)^2(2x - 5)^3[20x^2 - 30x + 8]$

We can factor out 2:

$= 2(x^2 + 1)^2(2x - 5)^3[10x^2 - 15x + 4]$

Answer: $g'(x) = 2(x^2 + 1)^2(2x - 5)^3(10x^2 - 15x + 4)$