Quotient Rule of Differentiation

First identify numerator f and denominator g. Form both derivatives f′ and g′. Then insert into $(\frac{f}{g})$′ = (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: $h(x) =\frac{ x^{2}}{(x + 1)}$ with f′ = 2x and g′ = 1 gives h′$(x) =\frac{ (2x(x + 1) - x^{2})}{(x + 1)^{2}} =\frac{ (x^{2} + 2x)}{(x + 1)^{2}}$. 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 $g(x) \ne{} 0$.

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: $(\frac{x}{x})$′ must give 0. Correct: $\frac{(1\cdot{}x - x\cdot{}1)}{x^{2}} = 0$ ✓. Swapping the order also gives 0 here, so better test with $(\frac{x^{2}}{x})$′: correct is $\frac{(2x\cdot{}x - x^{2}\cdot{}1)}{x^{2}} = 1$, as it must be for $\frac{x^{2}}{x} = x$; 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 $\frac{(2x + 1)}{(x - 3)}$ or $\frac{\sin{} x}{x}$. If the denominator is just a power, rewriting is faster: $\frac{3}{x^{2}} = 3x^{-2}$ differentiates by the power rule to $-6x^{-3}$, with no fraction work at all. If the numerator is just a constant, the reciprocal rule $(\frac{1}{g})$′ = −g′/g² suffices. And sometimes the fraction cancels before differentiating: $\frac{(x^{3} + x)}{x} = x^{2} + 1$ 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 $\tan{} x =\frac{ \sin{} x}{\cos{}} x$ and apply the quotient rule with $f = \sin{} x$, $g = \cos{} x$: f′ = cos x and g′ = −sin x. The numerator becomes $\cos{} x\cdot{}\cos{} x - \sin{} x\cdot{}(-\sin{} x) = \cos{}^{2}x + \sin{}^{2}x$, the denominator $\cos{}^{2}x$. With the trigonometric Pythagoras $\cos{}^{2}x + \sin{}^{2}x = 1$ it follows that (tan x)′ = 1/cos²x. Dividing numerator and denominator by $\cos{}^{2}x$ instead gives the equivalent form $1 + \tan{}^{2}x$. 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 $\frac{1}{\cos{}^{2}}x > 0$.