The Product Rule

In Words

First times derivative of second, PLUS second times derivative of first

Or think of it as: $[\text{first}]' \cdot [\text{second}] + [\text{first}] \cdot [\text{second}]'$

Why Not Just Multiply the Derivatives?

❌ Wrong: $(fg)' \neq f' \cdot g'$

This would miss information! The product rule is necessary.

Basic Example

Find $\frac{d}{dx}[(x^2)(x^3)]$

Method 1: Using Product Rule

Let $f(x) = x^2$ and $g(x) = x^3$

• $f'(x) = 2x$
• $g'(x) = 3x^2$

Apply the rule: $= f'(x) \cdot g(x) + f(x) \cdot g'(x)$ $= (2x)(x^3) + (x^2)(3x^2)$ $= 2x^4 + 3x^4 = 5x^4$

Method 2: Simplify First

$\frac{d}{dx}[(x^2)(x^3)] = \frac{d}{dx}[x^5] = 5x^4$ ✓

Same answer! But the product rule is needed when you can't simplify first.

Step-by-Step Process

Find: $\frac{d}{dx}[(3x + 1)(x^2 - 5)]$

Step 1: Identify the two functions

• First function: $f(x) = 3x + 1$
• Second function: $g(x) = x^2 - 5$

Step 2: Find their derivatives

• $f'(x) = 3$
• $g'(x) = 2x$

Step 3: Apply product rule $= f'(x) \cdot g(x) + f(x) \cdot g'(x)$ $= (3)(x^2 - 5) + (3x + 1)(2x)$

Step 4: Expand and simplify $= 3x^2 - 15 + 6x^2 + 2x$ $= 9x^2 + 2x - 15$

Memory Tricks

"First dee-second plus second dee-first"

Visual: Draw a cross:

  f   ·   g
   \   /
    \ /
     X
    / \
   /   \
  f'  ·  g  +  f  ·  g'

When to Use Product Rule

Use it when you have:

• Two functions multiplied: $(\text{stuff}) \times (\text{stuff})$
• Can't easily simplify first
• Each "stuff" needs its own derivative

Example with More Complex Functions

Find $\frac{d}{dx}[x^5(2x^2 + 3x)]$

Step 1: Identify

• $f(x) = x^5$, so $f'(x) = 5x^4$
• $g(x) = 2x^2 + 3x$, so $g'(x) = 4x + 3$

Step 2: Apply product rule $= (5x^4)(2x^2 + 3x) + (x^5)(4x + 3)$

Step 3: Expand $= 10x^6 + 15x^5 + 4x^6 + 3x^5$

Step 4: Combine like terms $= 14x^6 + 18x^5$

Three or More Functions

For three functions, apply the product rule twice:

$\frac{d}{dx}[f \cdot g \cdot h] = f'gh + fg'h + fgh'$

Pattern: Differentiate each function once, multiply by the others, then add.

Common Mistakes

❌ Forgetting the second term Don't write: $(fg)' = f'g$ (incomplete!)

❌ Mixing up the order Both terms are needed: $f'g + fg'$

❌ Not simplifying Always combine like terms at the end!

When NOT to Use It

If you can simplify first, do it!

Example: $\frac{d}{dx}[x^3 \cdot x^2]$

Easy way: $\frac{d}{dx}[x^5] = 5x^4$

Hard way: Product rule (unnecessary here)

Practice Strategy

1. Identify the two functions clearly
2. Find each derivative separately
3. Write out the product rule formula
4. Substitute carefully
5. Expand and simplify

Step 2: Find the derivatives

• $f'(x) = 2x$
• $g'(x) = 3x^2 - 2$

Step 3: Apply product rule $\frac{dy}{dx} = f'(x) \cdot g(x) + f(x) \cdot g'(x)$

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

Step 4: Expand $= 2x^4 - 4x^2 + 3x^4 - 2x^2 + 3x^2 - 2$

Step 5: Combine like terms $= 5x^4 - 3x^2 - 2$

Answer: dy/dx = 5x⁴ - 3x² - 2

Let $u = 3x^2 - 5$ and $v = 2x + 7$

$u' = 6x$ $v' = 2$

Apply product rule:

$f'(x) = (6x)(2x + 7) + (3x^2 - 5)(2)$

$= 12x^2 + 42x + 6x^2 - 10$

$= 18x^2 + 42x - 10$

Alternative: Expand first, then differentiate:

$f(x) = 6x^3 + 21x^2 - 10x - 35$

$f'(x) = 18x^2 + 42x - 10$ ✓

$\frac{dy}{dx} = (3x^2)(\sin x) + (x^3)(\cos x)$

$\frac{dy}{dx} = 3x^2 \sin x + x^3 \cos x$

Can factor if desired:

$\frac{dy}{dx} = x^2(3\sin x + x\cos x)$