The derivative of a quotient is not the derivative of the
numerator divided by the derivative of the denominator. The video
below shows this with an example. Instead, we have

The derivative of the quotient is not
the quotient of the derivatives.

We write, briefly, $\displaystyle\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^2}=\displaystyle\frac{\text{lo
de hi}-\text {hi de lo}}{\text{lolo}}$, where
hi=numerator, lo=denominator, and de=differentiate.
The important thing to remember here is that unlike the product
rule, where $f'g+fg'=fg'+f'g$ and the order doesn't matter,
$gf'-fg'\not = fg'-gf'$. Lo comes first!

The quotient rule can be derived from the product rule. If we
write $\displaystyle f(x) = g(x)\frac{f(x)}{g(x)}$, then the
product rule says that $$ f'(x) = \left ( g(x)
\cdot\frac{f(x)}{g(x)} \right )'; \quad\text{ i.e, }\quad f'(x)=
g'(x) \frac{f(x)}{g(x)} + g(x) \left ( \frac{f(x)}{g(x)} \right
)'. $$ Solving for $\left( \frac{f(x)}{g(x)} \right )'$ gives $$
\left ( \frac{f(x)}{g(x)} \right )' = \frac{f'(x) -
g'(x)\frac{f(x)}{g(x)}}{g(x)} = \frac{g(x) f'(x) - f(x)
g'(x)}{[g(x)]^2}.$$