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Quotient Rule

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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

$$\frac{d}{dx}\left[\tan x\right] = \frac{d}{dx}\left[\frac{\sin x}{\cos x}\right] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x$$

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:

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