The Quotient Rule: Derivatives of Fractions
The derivative of a fraction is not the fraction of the derivatives.
d/dx(x²/x) = d/dx(x) = 1. But (2x)/(1) = 2x. Those don't match.
Here's the unlock: when you divide, the denominator fights back. As the bottom grows, the whole fraction shrinks. That's why the quotient rule has a minus sign — the denominator's growth subtracts from the rate of change. And the bottom squared in the denominator? That's because the denominator's effect compounds: you're dividing by something that's itself changing.
The quotient rule is the product rule wearing a disguise: f/g = f·g⁻¹. But the disguise produces a specific pattern:
d/dx [f(x)/g(x)] = (f'(x)·g(x) - f(x)·g'(x)) / [g(x)]²
"Low d-high minus high d-low, over low squared."
The Formula
d/dx [f(x)/g(x)] = (f'(x)·g(x) - f(x)·g'(x)) / [g(x)]²
Breaking it down:
- Numerator: (derivative of top × bottom) - (top × derivative of bottom)
- Denominator: bottom squared
Note the subtraction. That negative sign is easy to forget.
Example: x/(x+1)
Let f(x) = x, g(x) = x + 1
f'(x) = 1 g'(x) = 1
d/dx [x/(x+1)] = (1·(x+1) - x·1) / (x+1)² = (x + 1 - x) / (x+1)² = 1 / (x+1)²
Example: sin(x)/x
Let f(x) = sin(x), g(x) = x
f'(x) = cos(x) g'(x) = 1
d/dx [sin(x)/x] = (cos(x)·x - sin(x)·1) / x² = (x·cos(x) - sin(x)) / x²
Example: (x² + 1)/(x - 1)
f(x) = x² + 1, f'(x) = 2x g(x) = x - 1, g'(x) = 1
d/dx = (2x·(x-1) - (x²+1)·1) / (x-1)² = (2x² - 2x - x² - 1) / (x-1)² = (x² - 2x - 1) / (x-1)²
Deriving It from Product Rule
f/g = f · g⁻¹
Using the product rule: d/dx [f · g⁻¹] = f' · g⁻¹ + f · d/dx(g⁻¹)
Using the chain rule on g⁻¹: d/dx(g⁻¹) = -g⁻² · g' = -g'/g²
Substituting: = f'/g + f·(-g'/g²) = f'/g - f·g'/g² = (f'·g - f·g') / g²
That's the quotient rule. It's product rule + chain rule, packaged conveniently.
When to Use It
Use the quotient rule when you have a genuine fraction with a variable denominator.
For 3x²/5, just write it as (3/5)x² and use the power rule. No quotient rule needed.
For (x² + 1)/(x - 3), quotient rule is the right tool.
Alternative: Rewrite as Product
Sometimes it's easier to rewrite:
d/dx [1/(x² + 1)] = d/dx [(x² + 1)⁻¹]
Using chain rule: = -1·(x² + 1)⁻² · 2x = -2x/(x² + 1)²
Same answer as quotient rule, different path. Choose whichever feels simpler.
The Minus Sign
The quotient rule has a minus sign; the product rule has a plus.
Memory trick: when dividing, things "subtract" from each other. The top and bottom are "fighting."
More precisely: increasing the denominator makes the fraction smaller, so g' contributes negatively.
Common Mistakes
Wrong order in numerator: Wrong: f·g' - f'·g Right: f'·g - f·g'
"Derivative of top" comes first.
Forgetting to square the bottom: Wrong: (f'g - fg')/g Right: (f'g - fg')/g²
Special Case: 1/g(x)
d/dx [1/g(x)] = (0·g - 1·g')/g² = -g'/g²
This is the reciprocal rule. Useful on its own:
d/dx [1/x] = -1/x² d/dx [1/cos(x)] = sin(x)/cos²(x) = tan(x)sec(x)
Longer Example: tan(x)
tan(x) = sin(x)/cos(x)
f(x) = sin(x), f'(x) = cos(x) g(x) = cos(x), g'(x) = -sin(x)
d/dx [tan(x)] = (cos(x)·cos(x) - sin(x)·(-sin(x))) / cos²(x) = (cos²(x) + sin²(x)) / cos²(x) = 1 / cos²(x) = sec²(x)
The Pythagorean identity cleans things up nicely.
Higher Quotients
For (f/g)/h = f/(g·h), you could:
- Apply quotient rule to f/(g·h)
- Apply quotient rule twice: first f/g, then that over h
Option 1 is usually cleaner.
The Core Insight
The quotient rule isn't a fundamentally new operation. It's what you get when you combine:
- Product rule: f/g = f · g⁻¹
- Chain rule: d/dx(g⁻¹) = -g'/g²
The formula packages this combination for convenience.
When the denominator increases, the fraction decreases — hence the minus sign in (f'g - fg'). When the denominator varies, its rate of change g' affects the fraction's rate of change through that negative contribution.
Low d-high minus high d-low, over low squared. That's the rhythm.
Part 6 of the Calculus Derivatives series.
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