| The Quotient Rule $$ \frac{d}{dx}\left( \frac{f(x)}{g(x)}\right) = \frac{g(x) f'(x) - f(x) g'(x)}{\left(g(x)\right)^2}.$$ |
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 $$
\left ( \frac{f(x)}{g(x)} \right )' = \frac{f'(x) -
g'(x)\frac{f(x)}{g(x)}}{g(x)} = \frac{g(x) f'(x) - f(x)
g'(x)}{[g(x)]^2}.$$