The Quotient Rule

Differentiating fractions and rational functions

The Quotient Rule

When you have one function divided by another, use the Quotient Rule!

The Rule

If y=f(x)g(x)y = \frac{f(x)}{g(x)}, then:

ddx[f(x)g(x)]=f(x)g(x)f(x)g(x)[g(x)]2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{[g(x)]^2}

In Words

Bottom times derivative of top, MINUS top times derivative of bottom, ALL OVER bottom squared

[bottom][top][top][bottom][bottom]2\frac{[\text{bottom}] \cdot [\text{top}]' - [\text{top}] \cdot [\text{bottom}]'}{[\text{bottom}]^2}

Memory Tricks

"Low dee-high minus high dee-low, over low-low"

Or sing it to a tune:

♪ Low dee-high,
  Minus high dee-low,
  All over low squared! ♪

Visual:

    f
   ---
    g

becomes

  g·f' - f·g'
  -----------
      g²

Basic Example

Find ddx[x2x+1]\frac{d}{dx}\left[\frac{x^2}{x + 1}\right]

Step 1: Identify top and bottom

  • Top (numerator): f(x)=x2f(x) = x^2, so f(x)=2xf'(x) = 2x
  • Bottom (denominator): g(x)=x+1g(x) = x + 1, so g(x)=1g'(x) = 1

Step 2: Apply quotient rule =g(x)f(x)f(x)g(x)[g(x)]2= \frac{g(x) \cdot f'(x) - f(x) \cdot g'(x)}{[g(x)]^2}

=(x+1)(2x)(x2)(1)(x+1)2= \frac{(x + 1)(2x) - (x^2)(1)}{(x + 1)^2}

Step 3: Expand numerator =2x2+2xx2(x+1)2= \frac{2x^2 + 2x - x^2}{(x + 1)^2}

Step 4: Simplify =x2+2x(x+1)2= \frac{x^2 + 2x}{(x + 1)^2}

We can factor if desired: =x(x+2)(x+1)2= \frac{x(x + 2)}{(x + 1)^2}

Important: Order Matters!

The quotient rule has a minus sign, so order is crucial!

gffgfggfg \cdot f' - f \cdot g' \quad \neq \quad f \cdot g' - g \cdot f'

Always: bottom times top prime minus top times bottom prime

When Can You Avoid the Quotient Rule?

Case 1: Constant in numerator

ddx[5x2]=ddx[5x2]=10x3=10x3\frac{d}{dx}\left[\frac{5}{x^2}\right] = \frac{d}{dx}[5x^{-2}] = -10x^{-3} = -\frac{10}{x^3}

Use negative exponent instead!

Case 2: Factorable expression

Sometimes you can simplify before differentiating.

Step-by-Step Example

Find ddx[3x+2x21]\frac{d}{dx}\left[\frac{3x + 2}{x^2 - 1}\right]

Step 1: Identify

  • Top: f(x)=3x+2f(x) = 3x + 2, so f(x)=3f'(x) = 3
  • Bottom: g(x)=x21g(x) = x^2 - 1, so g(x)=2xg'(x) = 2x

Step 2: Write the quotient rule =gffgg2= \frac{g \cdot f' - f \cdot g'}{g^2}

Step 3: Substitute =(x21)(3)(3x+2)(2x)(x21)2= \frac{(x^2 - 1)(3) - (3x + 2)(2x)}{(x^2 - 1)^2}

Step 4: Expand numerator =3x23(6x2+4x)(x21)2= \frac{3x^2 - 3 - (6x^2 + 4x)}{(x^2 - 1)^2}

=3x236x24x(x21)2= \frac{3x^2 - 3 - 6x^2 - 4x}{(x^2 - 1)^2}

Step 5: Combine like terms =3x24x3(x21)2= \frac{-3x^2 - 4x - 3}{(x^2 - 1)^2}

Or factor out -1: =(3x2+4x+3)(x21)2= \frac{-(3x^2 + 4x + 3)}{(x^2 - 1)^2}

Quotient Rule vs. Rewriting

Sometimes rewriting is easier:

Hard way (quotient rule): ddx[1x3]\frac{d}{dx}\left[\frac{1}{x^3}\right]

Easy way (power rule): ddx[x3]=3x4=3x4\frac{d}{dx}[x^{-3}] = -3x^{-4} = -\frac{3}{x^4}

If the numerator is constant or simple, consider rewriting!

Product Rule + Chain Rule Alternative

You can also rewrite quotients as products:

f(x)g(x)=f(x)[g(x)]1\frac{f(x)}{g(x)} = f(x) \cdot [g(x)]^{-1}

Then use product rule + chain rule. But quotient rule is usually faster!

Common Mistakes

Wrong sign: gf+fgg \cdot f' + f \cdot g' (should be minus!)

Forgetting to square bottom: ...g\frac{...}{g} (should be g2g^2!)

Wrong order: fggff \cdot g' - g \cdot f' (backwards!)

Correct: gffgg2\frac{g \cdot f' - f \cdot g'}{g^2}

Simplification Tips

After applying the quotient rule:

  1. Expand the numerator completely
  2. Combine like terms
  3. Factor if possible
  4. Don't expand (g(x))2(g(x))^2 unless necessary

Practice Strategy

  1. Circle the numerator (top)
  2. Box the denominator (bottom)
  3. Find both derivatives
  4. Write the quotient rule formula
  5. Substitute carefully (watch the minus!)
  6. Simplify the numerator
  7. Leave denominator factored (usually)

📚 Practice Problems

1Problem 1medium

Question:

Find the derivative of y = (x + 3)/(x - 2)

💡 Show Solution

Step 1: Identify top and bottom

  • Top: f(x)=x+3f(x) = x + 3, so f(x)=1f'(x) = 1
  • Bottom: g(x)=x2g(x) = x - 2, so g(x)=1g'(x) = 1

Step 2: Apply quotient rule dydx=gffgg2\frac{dy}{dx} = \frac{g \cdot f' - f \cdot g'}{g^2}

=(x2)(1)(x+3)(1)(x2)2= \frac{(x - 2)(1) - (x + 3)(1)}{(x - 2)^2}

Step 3: Expand numerator =x2x3(x2)2= \frac{x - 2 - x - 3}{(x - 2)^2}

Step 4: Simplify =5(x2)2= \frac{-5}{(x - 2)^2}

Answer: dy/dx = -5/(x - 2)²

2Problem 2medium

Question:

Find the derivative of f(x)=x2+3xx2f(x) = \frac{x^2 + 3x}{x - 2}.

💡 Show Solution

Solution:

Quotient rule: (uv)=uvuvv2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

Let u=x2+3xu = x^2 + 3x and v=x2v = x - 2

u=2x+3u' = 2x + 3 v=1v' = 1

f(x)=(2x+3)(x2)(x2+3x)(1)(x2)2f'(x) = \frac{(2x + 3)(x - 2) - (x^2 + 3x)(1)}{(x - 2)^2}

Expand numerator:

=2x24x+3x6x23x(x2)2= \frac{2x^2 - 4x + 3x - 6 - x^2 - 3x}{(x - 2)^2}

=2x2x24x+3x3x6(x2)2= \frac{2x^2 - x^2 - 4x + 3x - 3x - 6}{(x - 2)^2}

=x24x6(x2)2= \frac{x^2 - 4x - 6}{(x - 2)^2}

3Problem 3medium

Question:

Find the derivative of f(x)=x2+3xx2f(x) = \frac{x^2 + 3x}{x - 2}.

💡 Show Solution

Solution:

Quotient rule: (uv)=uvuvv2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

Let u=x2+3xu = x^2 + 3x and v=x2v = x - 2

u=2x+3u' = 2x + 3 v=1v' = 1

f(x)=(2x+3)(x2)(x2+3x)(1)(x2)2f'(x) = \frac{(2x + 3)(x - 2) - (x^2 + 3x)(1)}{(x - 2)^2}

Expand numerator:

=2x24x+3x6x23x(x2)2= \frac{2x^2 - 4x + 3x - 6 - x^2 - 3x}{(x - 2)^2}

=2x2x24x+3x3x6(x2)2= \frac{2x^2 - x^2 - 4x + 3x - 3x - 6}{(x - 2)^2}

=x24x6(x2)2= \frac{x^2 - 4x - 6}{(x - 2)^2}

4Problem 4hard

Question:

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

💡 Show Solution

Step 1: Identify

  • Top: u=2x23xu = 2x^2 - 3x, so u=4x3u' = 4x - 3
  • Bottom: v=x2+1v = x^2 + 1, so v=2xv' = 2x

Step 2: Quotient rule f(x)=vuuvv2f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2}

=(x2+1)(4x3)(2x23x)(2x)(x2+1)2= \frac{(x^2 + 1)(4x - 3) - (2x^2 - 3x)(2x)}{(x^2 + 1)^2}

Step 3: Expand first term (x2+1)(4x3)=4x33x2+4x3(x^2 + 1)(4x - 3) = 4x^3 - 3x^2 + 4x - 3

Step 4: Expand second term (2x23x)(2x)=4x36x2(2x^2 - 3x)(2x) = 4x^3 - 6x^2

Step 5: Subtract =4x33x2+4x3(4x36x2)(x2+1)2= \frac{4x^3 - 3x^2 + 4x - 3 - (4x^3 - 6x^2)}{(x^2 + 1)^2}

=4x33x2+4x34x3+6x2(x2+1)2= \frac{4x^3 - 3x^2 + 4x - 3 - 4x^3 + 6x^2}{(x^2 + 1)^2}

Step 6: Combine like terms =3x2+4x3(x2+1)2= \frac{3x^2 + 4x - 3}{(x^2 + 1)^2}

Answer: f'(x) = (3x² + 4x - 3)/(x² + 1)²

5Problem 5hard

Question:

Find dydx\frac{dy}{dx} if y=cosx1+sinxy = \frac{\cos x}{1 + \sin x}.

💡 Show Solution

Solution:

Quotient rule with u=cosxu = \cos x and v=1+sinxv = 1 + \sin x:

u=sinxu' = -\sin x v=cosxv' = \cos x

dydx=(sinx)(1+sinx)(cosx)(cosx)(1+sinx)2\frac{dy}{dx} = \frac{(-\sin x)(1 + \sin x) - (\cos x)(\cos x)}{(1 + \sin x)^2}

=sinxsin2xcos2x(1+sinx)2= \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2}

Use Pythagorean identity: sin2x+cos2x=1\sin^2 x + \cos^2 x = 1

=sinx1(1+sinx)2= \frac{-\sin x - 1}{(1 + \sin x)^2}

=(sinx+1)(1+sinx)2= \frac{-(\sin x + 1)}{(1 + \sin x)^2}

=11+sinx= \frac{-1}{1 + \sin x}

6Problem 6hard

Question:

Find dydx\frac{dy}{dx} if y=cosx1+sinxy = \frac{\cos x}{1 + \sin x}.

💡 Show Solution

Solution:

Quotient rule with u=cosxu = \cos x and v=1+sinxv = 1 + \sin x:

u=sinxu' = -\sin x v=cosxv' = \cos x

dydx=(sinx)(1+sinx)(cosx)(cosx)(1+sinx)2\frac{dy}{dx} = \frac{(-\sin x)(1 + \sin x) - (\cos x)(\cos x)}{(1 + \sin x)^2}

=sinxsin2xcos2x(1+sinx)2= \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2}

Use Pythagorean identity: sin2x+cos2x=1\sin^2 x + \cos^2 x = 1

=sinx1(1+sinx)2= \frac{-\sin x - 1}{(1 + \sin x)^2}

=(sinx+1)(1+sinx)2= \frac{-(\sin x + 1)}{(1 + \sin x)^2}

=11+sinx= \frac{-1}{1 + \sin x}