Quotient Rule

(f/g)' = (f'g - fg') / g^2 — low d-high minus high d-low, over the square of the low.

The idea

Theorem (Quotient rule).

If $f$ and $g$ are differentiable and $g \neq 0$, then $\left(\frac{f}{g}\right)' = \frac{f'\,g - f\,g'}{g^{2}},$ where $f$ is the top of the fraction and $g$ the bottom.

The derivative of a quotient is not the quotient of the derivatives. The quotient $\frac{x^{2}}{x}$ equals $x$, whose derivative is $1$, while dividing the two derivatives would give $\frac{2x}{1} = 2x$.

In the numerator of the rule the order matters: the term carrying the differentiated top, $f'g$, comes first, and swapping the two terms flips the sign of the whole answer. The division is by $g^{2}$, not by $g$. And because $g^{2}$ sits in the denominator, the rule says nothing wherever $g = 0$ — exactly the points where the quotient itself is undefined.

Ways to work on it

• Walkthrough. Apply the quotient rule to a simple rational function at a point.
• Practice. Differentiate a rational function at a point with the quotient rule.
• Hardest. Differentiate a quotient symbolically and simplify the result.

Not sure where to start? Take the ten-question placement test.