For a fraction of two functions, $f$ on top and $g$ below, the quotient rule is $\left(\dfrac{f}{g}\right)' = \dfrac{f'g - fg'}{g^2}$, read aloud: the derivative of f over g is f prime g minus f g prime, all over g squared. The two terms on top are the product rule's two terms with the sign between them changed, and the bottom is squared. The order on top matters: the derivative of the TOP times the bottom leads, and the top times the derivative of the bottom is subtracted. Reverse them and every answer comes out with the wrong sign, which is the first trap in the lesson's name; forget the square and the answer is off by a factor of $g$, which is the second.

For $f(x) = \dfrac{x^2}{x + 1}$ at $x = 1$: $f = 1, f' = 2, g = 2, g' = 1$, so the slope is $\dfrac{2 \cdot 2 - 1 \cdot 1}{2^2} = \dfrac{3}{4} = 0.75$. Why the minus: a bigger bottom makes the fraction SMALLER, so the bottom's growth pulls the quotient down. Why the square: units guide. If $f$ is in metres and $g$ in seconds, $f/g$ is in m/s and its derivative with respect to a time $x$ in seconds is in m/s². The numerator $f'g - fg'$ is in m·s/s = m; over $g^2$ in s² that is m/s², which passes; over $g$ alone it would be m/s, which fails, no appeal. The failing unit proves $g^2$ belongs there; the passing unit proves nothing else.