Quotient Rule
The Formula
If $f$ and $g$ are differentiable, then:
$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}$$
Mnemonic
“Lo d-Hi minus Hi d-Lo over Lo squared”
- Lo = denominator $g(x)$
- Hi = numerator $f(x)$
- d = derivative
$$\frac{d}{dx}\left[\frac{\text{Hi}}{\text{Lo}}\right] = \frac{\text{Lo} \cdot d\text{Hi} - \text{Hi} \cdot d\text{Lo}}{(\text{Lo})^2}$$
Examples
Example 1
$$\frac{d}{dx}\left[\frac{x^2}{x+1}\right] = \frac{(x+1)(2x) - x^2(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}$$
Example 2
$$\frac{d}{dx}\left[\tan x\right] = \frac{d}{dx}\left[\frac{\sin x}{\cos x}\right] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x$$
the derivative of a quotient is the quotient of the derivatives.
This is the multiplicative-not-additive error. Just as $(fg)' \neq f'g'$, it is also false that $\left(\frac{f}{g}\right)' = \frac{f'}{g'}$. For $f(x) = x^2$ and $g(x) = x$: $\frac{f}{g} = x$, so $\left(\frac{f}{g}\right)' = 1$. But $\frac{f'}{g'} = \frac{2x}{1} = 2x \neq 1$. The correct Quotient Rule gives $\frac{g f' - f g'}{g^2} = \frac{x \cdot 2x - x^2 \cdot 1}{x^2} = \frac{x^2}{x^2} = 1$.
a large denominator makes the derivative large.
This is the height-vs-slope error. The derivative $\left(\frac{f}{g}\right)' = \frac{gf' - fg'}{g^2}$ shows the square of $g$ in the denominator. A large $g$ makes the derivative SMALLER, not larger. For $f(x) = 1$ and $g(x) = x^n$ (large $n$): $\frac{d}{dx}\left(\frac{1}{x^n}\right) = \frac{-n}{x^{n+1}}$. As $x$ grows, the function $1/x^n$ shrinks toward zero AND its derivative $-n/x^{n+1}$ also shrinks toward zero -- a flatter and flatter graph. The denominator $g^2$ in the Quotient Rule acts as a damping factor, not an amplifier.
Common Mistakes
- Wrong order: The formula is NOT $\frac{f'g - fg'}{g^2}$; it’s $\frac{gf' - fg'}{g^2}$
- Forgetting to square: The denominator is $g^2$, not $g$
- Sign error: There’s a minus sign between the two terms
When to Use
- Differentiating rational functions
- Deriving trig identities ($\tan x = \frac{\sin x}{\cos x}$, etc.)
- Any quotient of differentiable functions
Alternative: Negative Exponent
Sometimes it’s easier to rewrite $\frac{f}{g} = f \cdot g^{-1}$ and use the product rule with chain rule:
$$\frac{d}{dx}[f \cdot g^{-1}] = f' \cdot g^{-1} + f \cdot (-1)g^{-2}g'$$