Why is the quotient rule in differentiation necessary? 
*

*Calculus - Derivatives - Quotient Rule
Why is a quotient rule even necessary?
Why can't we just consider $\frac{A}{B}$ as $A \cdot B^{-1}$ and use the multiplication formula?
 A: Why to have a rule for the derivative of $A^3$ if we can write it as $A\cdot (A\cdot A)$ and apply two times the product rule?
A: Logically the quotient rule in calculus is not needed, since it can be derived from the product rule, the power rule, and the chain rule every time, e.g., $(1/g)' = (g^{-1})' = -g^{-2}g' = -g'/g^2$. But most students learn the quotient rule and don't have trouble after practicing it(and then they have to learn not to confuse it with the ratio of derivatives in L'Hopital's rule later).
I once met a very famous mathematician who does not  know the quotient rule: he learned math in Europe, where university courses often begin with analysis rather than elementary calculus, and he never teaches freshman calculus so he has no reason to make contact with the quotient rule. I was discussing something with him and when the derivative of a ratio was needed he found it with the product rule and told me he didn't know another way and didn't care if there is another way.
A: I like using
$(\ln(f))'
=\dfrac{f'}{f}
$
for general
products and quotients.
If
$f = \dfrac{\prod u_k}{\prod v_k}
$
then
$\ln(f) 
= \sum \ln(u_k)-\sum \ln(v_k)
$
so
$(\ln(f))' 
= \sum \dfrac{u_k'}{u_k}-\sum \dfrac{v_k'}{v_k}
=\dfrac{f'}{f}
$
so
$f'
= f\left(\sum \dfrac{u_k'}{u_k}-\sum \dfrac{v_k'}{v_k}\right)
$.
From this
all the product and quotient rules
are special cases.
For example,
if $f = \dfrac{u}{v}
$
then
$\begin{array}\\
f'
&=f(\dfrac{u'}{u}-\dfrac{v'}{v})\\
&=\dfrac{u}{v}(\dfrac{u'v-v'u}{uv})\\
&=\dfrac{u'v-v'u}{v^2}\\
\end{array}
$
Since this is formal,
I don't worry about the sign
of the $u_k$ and $v_k$.
A: You can derive the quotient rule by considering $\dfrac{f(x)}{g(x)}$ as $f(x)\cdot(g(x))^{-1}$ and then using the product and chain rule.The quotient rule  gives a formula (under the right conditions) for evaluating  the derivative of  a quotient without using the product and chain rule each time.
A: Personally, I have never used the quotient rule as I have always found it ugly and inconvenient. For example try computing the derivative of $\large\frac{(x+1)^3}{(2x-4)^5}$ with respect to $x$. Since the quotient rule can be derived from the product rule and chain rule, it is indeed redundant if you have the other two as well as the derivative of $x^{-1}$ with respect to $x$.
One might point out that we could just as well say that all the rules are redundant if you have the definition of derivative. However, there is something called the formal derivative that is defined algebraically and not via a limit-based definition (because it makes no sense in that context). So there is some benefit (beyond just for ease of computation) in finding purely algebraic rules for differentiation.
