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The Quotient Rule

Master the technique for differentiating ratios of functions: when one function is divided by another, the Quotient Rule gives you the derivative.

AP Calculus BC

Why Do We Need a Quotient Rule?

Many real-world functions are expressed as one quantity divided by another: rates, concentrations, probabilities, and ratios. Simply differentiating the numerator and denominator separately does not give the correct derivative.

The Building Blocks

\( f(x) \): Numerator
The function on top of the fraction. Often called "hi" in the mnemonic.
\( g(x) \): Denominator
The function on the bottom. Often called "lo." Must satisfy \( g(x) \neq 0 \).
\( h(x) \): The Quotient
The ratio \( h(x) = \dfrac{f(x)}{g(x)} \) whose derivative we seek.

From First Principles to the Rule

We can derive the Quotient Rule using the limit definition of the derivative:

$$ h'(x) = \lim_{\Delta x \to 0} \frac{\dfrac{f(x+\Delta x)}{g(x+\Delta x)} - \dfrac{f(x)}{g(x)}}{\Delta x} $$

By combining the fractions in the numerator over a common denominator, expanding, and taking the limit, we arrive at the Quotient Rule formula.

The Quotient Rule Formula

$$ \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{\bigl[g(x)\bigr]^2} $$
\( f'(x) \) · \( g(x) \) − \( f(x) \) · \( g'(x) \)
\( [\,g(x)\,]^2 \)
NUMERATOR DENOMINATOR

Connection to the Product Rule

You can also view division as multiplication by a reciprocal. Since \( \dfrac{f}{g} = f \cdot g^{-1} \), applying the Product Rule and Chain Rule yields the same result:

$$ \frac{d}{dx}\bigl[f \cdot g^{-1}\bigr] = f'\,g^{-1} + f\,(-1)\,g^{-2}\,g' = \frac{f'\,g - f\,g'}{g^2} $$

Quotient vs Product Rule

Quotient Rule Product Rule
Form \( \dfrac{f}{g} \) \( f \cdot g \)
Formula \( \dfrac{f'g - fg'}{g^2} \) \( f'g + fg' \)
Sign Subtraction (order matters) Addition (order irrelevant)
Denominator \( g^2 \) required None
Key Pitfall Reversing \( f'g \) and \( fg' \) Forgetting to differentiate both

Step-by-Step Application

1
Identify \( f(x) \) and \( g(x) \). The numerator is \( f \); the denominator is \( g \).
2
Differentiate each separately. Find \( f'(x) \) and \( g'(x) \) using basic rules.
3
Plug into the formula. Compute \( f'g - fg' \) for the numerator, \( g^2 \) for the denominator.
4
Expand and simplify. Distribute, combine like terms, and factor if possible.
5
State domain restrictions. The derivative is undefined wherever \( g(x) = 0 \).

Example 1: Polynomial Quotient

Find the derivative of \( h(x) = \dfrac{x^2 + 1}{x - 3} \).

1
Identify: \( f(x) = x^2 + 1 \), \( g(x) = x - 3 \).
2
Differentiate: \( f'(x) = 2x \), \( g'(x) = 1 \).
3
Apply formula: \( h'(x) = \dfrac{(2x)(x-3) - (x^2+1)(1)}{(x-3)^2} \).
4
Simplify: \( = \dfrac{2x^2 - 6x - x^2 - 1}{(x-3)^2} = \dfrac{x^2 - 6x - 1}{(x-3)^2} \).

Example 2: Trigonometric Quotient

Derive \( \dfrac{d}{dx}[\tan x] \) using the Quotient Rule on \( \dfrac{\sin x}{\cos x} \).

1
Set up: \( f(x) = \sin x \), \( g(x) = \cos x \).
2
Differentiate: \( f'(x) = \cos x \), \( g'(x) = -\sin x \).
3
Apply: \( \dfrac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} \).
4
Simplify: By the Pythagorean identity, \( = \dfrac{1}{\cos^2 x} = \sec^2 x \).

Example 3: Exponential Quotient

Find \( h'(x) \) for \( h(x) = \dfrac{e^x}{x^2 + 1} \).

1
Identify: \( f(x) = e^x \), \( g(x) = x^2 + 1 \).
2
Differentiate: \( f'(x) = e^x \), \( g'(x) = 2x \).
3
Apply: \( h'(x) = \dfrac{e^x(x^2+1) - e^x(2x)}{(x^2+1)^2} \).
4
Factor: \( = \dfrac{e^x(x^2 - 2x + 1)}{(x^2+1)^2} = \dfrac{e^x(x-1)^2}{(x^2+1)^2} \).

Visualizing \( \dfrac{e^x}{x^2+1} \)

-3 -2 -1 0 1 2 3 1 2

When Not to Use the Quotient Rule

Sometimes rewriting the expression avoids the Quotient Rule entirely, making differentiation simpler and faster.

Constant Denominator
\( \dfrac{x^3}{5} = \dfrac{1}{5}x^3 \). Just use the Constant Multiple Rule.
Monomial Denominator
\( \dfrac{x^3 + 2x}{x} = x^2 + 2 \). Simplify first, then use the Power Rule.
Negative Exponent
\( \dfrac{1}{x^n} = x^{-n} \). The Power Rule handles this directly.

Trig Derivatives via the Quotient Rule

All six trig derivatives can be found. The Quotient Rule derives three more beyond the basic \( \sin \) and \( \cos \):

Function Written As Derivative
\( \tan x \) \( \dfrac{\sin x}{\cos x} \) \( \sec^2 x \)
\( \cot x \) \( \dfrac{\cos x}{\sin x} \) \( -\csc^2 x \)
\( \sec x \) \( \dfrac{1}{\cos x} \) \( \sec x \tan x \)
\( \csc x \) \( \dfrac{1}{\sin x} \) \( -\csc x \cot x \)

Common Mistakes to Avoid

✕ Swapping the Order
Incorrectly writing the numerator as \( fg' - f'g \). The correct order is \( f'g - fg' \). The subtraction is not commutative.
✕ Forgetting \( g^2 \)
Omitting the denominator \( [g(x)]^2 \). The entire expression must be divided by the square of the original denominator.
✕ Sign Errors
Dropping the negative sign when \( g'(x) \) is negative. Be careful with chain-rule derivatives inside \( g'(x) \).

Summary and Key Takeaways

1
The Quotient Rule: \( \dfrac{d}{dx}\!\left[\dfrac{f}{g}\right] = \dfrac{f'g - fg'}{g^2} \).
2
Order matters: it is \( f'g - fg' \), never the reverse.
3
The rule follows from the Product Rule applied to \( f \cdot g^{-1} \).
4
Simplify first whenever possible: constants, monomials, and negative exponents don't need it.
5
The Quotient Rule derives \( \tan x \), \( \cot x \), \( \sec x \), and \( \csc x \) derivatives.

Question 1 of 5

What is \( \dfrac{d}{dx}\!\left[\dfrac{x}{x+1}\right] \)?

Question 2 of 5

Using the Quotient Rule, what is \( \dfrac{d}{dx}[\cot x] \)?

Question 3 of 5

If \( h(x) = \dfrac{e^x}{x^2+1} \), what is the value of \( h'(0) \)?

Question 4 of 5

In the Quotient Rule formula \( \dfrac{f'g - fg'}{g^2} \), what happens if you swap the order to \( fg' - f'g \)?

Question 5 of 5

Find \( \dfrac{d}{dx}\!\left[\dfrac{\ln x}{x}\right] \) for \( x > 0 \).