The Product Rule

Differentiating the product of two functions

The Product Rule

When you have two functions multiplied together, you need the Product Rule!

The Rule

If y=f(x)g(x)y = f(x) \cdot g(x), then:

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)\frac{d}{dx}[f(x) \cdot g(x)] = f'(x) \cdot g(x) + f(x) \cdot g'(x)

In Words

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

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

Why Not Just Multiply the Derivatives?

Wrong: (fg)fg(fg)' \neq f' \cdot g'

This would miss information! The product rule is necessary.

Basic Example

Find ddx[(x2)(x3)]\frac{d}{dx}[(x^2)(x^3)]

Method 1: Using Product Rule

Let f(x)=x2f(x) = x^2 and g(x)=x3g(x) = x^3

  • f(x)=2xf'(x) = 2x
  • g(x)=3x2g'(x) = 3x^2

Apply the rule: =f(x)g(x)+f(x)g(x)= f'(x) \cdot g(x) + f(x) \cdot g'(x) =(2x)(x3)+(x2)(3x2)= (2x)(x^3) + (x^2)(3x^2) =2x4+3x4=5x4= 2x^4 + 3x^4 = 5x^4

Method 2: Simplify First

ddx[(x2)(x3)]=ddx[x5]=5x4\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: ddx[(3x+1)(x25)]\frac{d}{dx}[(3x + 1)(x^2 - 5)]

Step 1: Identify the two functions

  • First function: f(x)=3x+1f(x) = 3x + 1
  • Second function: g(x)=x25g(x) = x^2 - 5

Step 2: Find their derivatives

  • f(x)=3f'(x) = 3
  • g(x)=2xg'(x) = 2x

Step 3: Apply product rule =f(x)g(x)+f(x)g(x)= f'(x) \cdot g(x) + f(x) \cdot g'(x) =(3)(x25)+(3x+1)(2x)= (3)(x^2 - 5) + (3x + 1)(2x)

Step 4: Expand and simplify =3x215+6x2+2x= 3x^2 - 15 + 6x^2 + 2x =9x2+2x15= 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: (stuff)×(stuff)(\text{stuff}) \times (\text{stuff})
  • Can't easily simplify first
  • Each "stuff" needs its own derivative

Example with More Complex Functions

Find ddx[x5(2x2+3x)]\frac{d}{dx}[x^5(2x^2 + 3x)]

Step 1: Identify

  • f(x)=x5f(x) = x^5, so f(x)=5x4f'(x) = 5x^4
  • g(x)=2x2+3xg(x) = 2x^2 + 3x, so g(x)=4x+3g'(x) = 4x + 3

Step 2: Apply product rule =(5x4)(2x2+3x)+(x5)(4x+3)= (5x^4)(2x^2 + 3x) + (x^5)(4x + 3)

Step 3: Expand =10x6+15x5+4x6+3x5= 10x^6 + 15x^5 + 4x^6 + 3x^5

Step 4: Combine like terms =14x6+18x5= 14x^6 + 18x^5

Three or More Functions

For three functions, apply the product rule twice:

ddx[fgh]=fgh+fgh+fgh\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)=fg(fg)' = f'g (incomplete!)

Mixing up the order Both terms are needed: fg+fgf'g + fg'

Not simplifying Always combine like terms at the end!

When NOT to Use It

If you can simplify first, do it!

Example: ddx[x3x2]\frac{d}{dx}[x^3 \cdot x^2]

Easy way: ddx[x5]=5x4\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

📚 Practice Problems

1Problem 1medium

Question:

Find the derivative of y = (x² + 1)(x³ - 2x)

💡 Show Solution

Step 1: Identify the functions

  • First: f(x)=x2+1f(x) = x^2 + 1
  • Second: g(x)=x32xg(x) = x^3 - 2x

Step 2: Find the derivatives

  • f(x)=2xf'(x) = 2x
  • g(x)=3x22g'(x) = 3x^2 - 2

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

=(2x)(x32x)+(x2+1)(3x22)= (2x)(x^3 - 2x) + (x^2 + 1)(3x^2 - 2)

Step 4: Expand =2x44x2+3x42x2+3x22= 2x^4 - 4x^2 + 3x^4 - 2x^2 + 3x^2 - 2

Step 5: Combine like terms =5x43x22= 5x^4 - 3x^2 - 2

Answer: dy/dx = 5x⁴ - 3x² - 2

2Problem 2medium

Question:

Find the derivative of f(x)=(3x25)(2x+7)f(x) = (3x^2 - 5)(2x + 7).

💡 Show Solution

Solution:

Product rule: (uv)=uv+uv(uv)' = u'v + uv'

Let u=3x25u = 3x^2 - 5 and v=2x+7v = 2x + 7

u=6xu' = 6x v=2v' = 2

Apply product rule:

f(x)=(6x)(2x+7)+(3x25)(2)f'(x) = (6x)(2x + 7) + (3x^2 - 5)(2)

=12x2+42x+6x210= 12x^2 + 42x + 6x^2 - 10

=18x2+42x10= 18x^2 + 42x - 10

Alternative: Expand first, then differentiate:

f(x)=6x3+21x210x35f(x) = 6x^3 + 21x^2 - 10x - 35

f(x)=18x2+42x10f'(x) = 18x^2 + 42x - 10

3Problem 3medium

Question:

Find the derivative of f(x)=(3x25)(2x+7)f(x) = (3x^2 - 5)(2x + 7).

💡 Show Solution

Solution:

Product rule: (uv)=uv+uv(uv)' = u'v + uv'

Let u=3x25u = 3x^2 - 5 and v=2x+7v = 2x + 7

u=6xu' = 6x v=2v' = 2

Apply product rule:

f(x)=(6x)(2x+7)+(3x25)(2)f'(x) = (6x)(2x + 7) + (3x^2 - 5)(2)

=12x2+42x+6x210= 12x^2 + 42x + 6x^2 - 10

=18x2+42x10= 18x^2 + 42x - 10

Alternative: Expand first, then differentiate:

f(x)=6x3+21x210x35f(x) = 6x^3 + 21x^2 - 10x - 35

f(x)=18x2+42x10f'(x) = 18x^2 + 42x - 10

4Problem 4hard

Question:

Find f'(x) if f(x) = (2x - 5)(3x² + x - 1)

💡 Show Solution

Step 1: Identify

  • First: u=2x5u = 2x - 5, so u=2u' = 2
  • Second: v=3x2+x1v = 3x^2 + x - 1, so v=6x+1v' = 6x + 1

Step 2: Product rule f(x)=uv+uvf'(x) = u'v + uv' =(2)(3x2+x1)+(2x5)(6x+1)= (2)(3x^2 + x - 1) + (2x - 5)(6x + 1)

Step 3: Expand first term 2(3x2+x1)=6x2+2x22(3x^2 + x - 1) = 6x^2 + 2x - 2

Step 4: Expand second term (FOIL) (2x5)(6x+1)=12x2+2x30x5=12x228x5(2x - 5)(6x + 1) = 12x^2 + 2x - 30x - 5 = 12x^2 - 28x - 5

Step 5: Add them together f(x)=6x2+2x2+12x228x5f'(x) = 6x^2 + 2x - 2 + 12x^2 - 28x - 5 =18x226x7= 18x^2 - 26x - 7

Answer: f'(x) = 18x² - 26x - 7

5Problem 5hard

Question:

Find dydx\frac{dy}{dx} if y=x3sinxy = x^3 \sin x.

💡 Show Solution

Solution:

Product rule with u=x3u = x^3 and v=sinxv = \sin x:

u=3x2u' = 3x^2 v=cosxv' = \cos x

dydx=(3x2)(sinx)+(x3)(cosx)\frac{dy}{dx} = (3x^2)(\sin x) + (x^3)(\cos x)

dydx=3x2sinx+x3cosx\frac{dy}{dx} = 3x^2 \sin x + x^3 \cos x

Can factor if desired:

dydx=x2(3sinx+xcosx)\frac{dy}{dx} = x^2(3\sin x + x\cos x)

6Problem 6hard

Question:

Find dydx\frac{dy}{dx} if y=x3sinxy = x^3 \sin x.

💡 Show Solution

Solution:

Product rule with u=x3u = x^3 and v=sinxv = \sin x:

u=3x2u' = 3x^2 v=cosxv' = \cos x

dydx=(3x2)(sinx)+(x3)(cosx)\frac{dy}{dx} = (3x^2)(\sin x) + (x^3)(\cos x)

dydx=3x2sinx+x3cosx\frac{dy}{dx} = 3x^2 \sin x + x^3 \cos x

Can factor if desired:

dydx=x2(3sinx+xcosx)\frac{dy}{dx} = x^2(3\sin x + x\cos x)