Quotient Rule of Differentiation

First identify numerator f and denominator g. Form both derivatives f′ and g′. Then insert into (f/g)′ = (f′g − fg′)/g²: derivative of the numerator times denominator, minus numerator times derivative of the denominator, all over the square of the denominator. Example: h(x) = x²/(x + 1) with f′ = 2x and g′ = 1 gives h′(x) = (2x(x + 1) − x²)/(x + 1)² = (x² + 2x)/(x + 1)². Simplify at the end, but usually leave the denominator factored, which makes zero and domain questions easier. Important: the order in the numerator cannot be swapped because of the minus sign, and the rule holds only where g(x) ≠ 0.

A common English mnemonic is "low d-high minus high d-low, over the square of what is below": denominator times derivative of the numerator, minus numerator times derivative of the denominator, over the denominator squared. Both phrasings describe the same expression (f′g − fg′)/g². The decisive point: the term with the DIFFERENTIATED numerator stands first and positive. A quick self-test exposes mix-ups: (x/x)′ must give 0. Correct: (1·x − x·1)/x² = 0 ✓. Swapping the order also gives 0 here, so better test with (x²/x)′: correct is (2x·x − x²·1)/x² = 1, as it must be for x²/x = x; the swapped version would give −1. This unmasks the error in seconds.

The quotient rule pays off when numerator and denominator are both genuine functions of x, for example (2x + 1)/(x − 3) or sin x/x. If the denominator is just a power, rewriting is faster: 3/x² = 3x⁻² differentiates by the power rule to −6x⁻³, with no fraction work at all. If the numerator is just a constant, the reciprocal rule (1/g)′ = −g′/g² suffices. And sometimes the fraction cancels before differentiating: (x³ + x)/x = x² + 1 is trivial to differentiate. Rule of thumb: simplify first, then differentiate. Reaching reflexively for the quotient rule produces longer terms and more error sources than necessary.

Four classics. First: numerator order swapped, i.e. fg′ − f′g instead of f′g − fg′; this flips the sign of the entire derivative. Second: forgetting the square in the denominator and dividing only by g. Third: putting a plus as in the product rule; the product rule has plus, the quotient rule minus. Fourth: cancelling wrongly after differentiating, for instance individual summands of the numerator against the denominator; you may only cancel common factors of the whole numerator. Antidote: write the formula out cleanly, note f, g, f′, g′ separately, only then substitute, and check the result numerically at a simple point like x = 0 or x = 1 against the difference quotient.

Write tan x = sin x/cos x and apply the quotient rule with f = sin x, g = cos x: f′ = cos x and g′ = −sin x. The numerator becomes cos x·cos x − sin x·(−sin x) = cos²x + sin²x, the denominator cos²x. With the trigonometric Pythagoras cos²x + sin²x = 1 it follows that (tan x)′ = 1/cos²x. Dividing numerator and denominator by cos²x instead gives the equivalent form 1 + tan²x. This derivation is a popular exam transfer task, because it combines the quotient rule, the derivatives of the trigonometric functions and the trigonometric identity in three lines. The result also shows: tan x increases everywhere on its branches, since 1/cos²x > 0.