The Chain Rule

Finding derivatives of composite functions

🔗 The Chain Rule

What is the Chain Rule?

The Chain Rule is one of the most important and widely used differentiation rules. It tells us how to find the derivative of a composite function (a function inside another function).

📊 The Formula

If $y = f(g(x))$ , then:

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

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

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

The "Outside-Inside" Method

Step-by-Step Process

Identify the outside function and the inside function
Differentiate the outside function (leave the inside alone)
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:

Outside: squaring function
Middle: sine function
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

Circle the inside function
Box the outside function
Write "outside' × inside'" as a reminder
Apply the formula step by step
Simplify your answer

📚 Practice Problems

1Problem 1medium

❓ Question:

Find the derivative of each function:

a) $f(x) = (3x^2 - 5)^7$ b) $g(x) = \sin(4x)$ c) $h(x) = e^{x^2 + 1}$

💡 Show Solution

Solution:

Part (a): Chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

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 $y = e^u$

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

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

2Problem 2medium

❓ Question:

Find the derivative of $f(x) = (2x^3 - 5x + 1)^6$ using the Chain Rule.

💡 Show Solution

Step 1: Identify the functions

Outside function: $u^6$ (something to the 6th power)

Inside function: $u = 2x^3 - 5x + 1$

Step 2: Apply the Chain Rule

Using $\frac{d}{dx}[u^6] = 6u^5 \cdot \frac{du}{dx}$

Step 3: Find the derivative of the inside

$\frac{du}{dx} = \frac{d}{dx}[2x^3 - 5x + 1] = 6x^2 - 5$

Step 4: Put it together

$f'(x) = 6(2x^3 - 5x + 1)^5 \cdot (6x^2 - 5)$

This can also be written as:

$f'(x) = 6(6x^2 - 5)(2x^3 - 5x + 1)^5$

Answer: $f'(x) = 6(6x^2 - 5)(2x^3 - 5x + 1)^5$

3Problem 3medium

❓ Question:

Find the derivative of each function:

a) $f(x) = (3x^2 - 5)^7$ b) $g(x) = \sin(4x)$ c) $h(x) = e^{x^2 + 1}$

💡 Show Solution

Solution:

Part (a): Chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

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 $y = e^u$

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

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

4Problem 4hard

❓ Question:

Find $\frac{dy}{dx}$ if $y = \sqrt{\cos(3x^2)}$ .

💡 Show Solution

Solution:

Rewrite: $y = [\cos(3x^2)]^{1/2}$

This requires chain rule twice (nested composition).

Let $u = \cos(3x^2)$ , then $y = u^{1/2}$

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

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:

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

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

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

5Problem 5hard

❓ Question:

Find $\frac{dy}{dx}$ if $y = \sqrt{\cos(3x^2)}$ .

💡 Show Solution

Solution:

Rewrite: $y = [\cos(3x^2)]^{1/2}$

This requires chain rule twice (nested composition).

Let $u = \cos(3x^2)$ , then $y = u^{1/2}$

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

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:

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

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

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

6Problem 6medium

❓ Question:

Find $\frac{dy}{dx}$ if $y = \sqrt{4x^2 + 9}$ .

💡 Show Solution

Step 1: Rewrite using exponents

$y = \sqrt{4x^2 + 9} = (4x^2 + 9)^{1/2}$

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

7Problem 7hard

❓ Question:

Find the derivative of $g(x) = (x^2 + 1)^3(2x - 5)^4$ . (This requires both Product Rule AND Chain Rule!)

💡 Show Solution

This problem requires both the Product Rule and the Chain Rule.

Step 1: Identify that this is a product

Let $u = (x^2 + 1)^3$ and $v = (2x - 5)^4$

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

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