4
$\begingroup$

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?

$\endgroup$
11
  • $\begingroup$ How do you find $\frac{d}{dx}\frac{1}{B}$ then? $\endgroup$ May 21, 2021 at 17:08
  • 14
    $\begingroup$ the general quotient rule for $\frac{f}{g}$ can be deduced as a corollary of the product rule, chain rule, and the rule for differentiating $f(x)=\frac{1}{x}$. So, strictly speaking, there's no need to make rule out of it, but the thing is quotients appear so often that is is worthwhile just knowing how to differentiate quotients directly just by looking at them. $\endgroup$
    – peek-a-boo
    May 21, 2021 at 17:09
  • 1
    $\begingroup$ How do you think the quotient rule is derived? $\endgroup$ May 21, 2021 at 17:12
  • 7
    $\begingroup$ @JoshuaWang $(B)^{-1}$ is clearly intended here as the reciprocal, not $B^{(-1)}$ the inverse. This conflation of notation is a scourge to math educators, which is why $\arcsin$ is used. And OP has it correct that it is sufficient to derive the quotient rule therefrom. I do it all the time if I can't remember which part gets subtracted. $\endgroup$
    – obscurans
    May 21, 2021 at 17:17
  • 3
    $\begingroup$ @JoshuaWang Although it can be ambiguous, and some people don't like it, the notation $f^{-1}(x)$ is widely used for both $1/f(x)$ and the inverse function of $f$, usually been clear from the context which meaning is been used. In this case, is clear from the context that it means the reciprocal. $\endgroup$
    – jjagmath
    May 21, 2021 at 17:37

5 Answers 5

10
$\begingroup$

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 derivatives of a quotient without using the product and chain rule each time.

$\endgroup$
1
  • 6
    $\begingroup$ IOW it's not necessary, just a shortcut result $\endgroup$
    – obscurans
    May 21, 2021 at 17:18
6
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ This should be a comment, not an answer. $\endgroup$
    – Tavish
    Jun 15, 2021 at 12:46
2
$\begingroup$

Logically the quotient rule in analysis is not needed, since it comes from the product rule, the rule for differentiating inverses, and the chain rule, e.g., $(1/g)' = (g^{-1})' = -g^{-2}g' = -g'/g^2$. Most students learn it anyway (and then 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.

$\endgroup$
2
$\begingroup$

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 $v_k$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ There is a formal derivative of $x^{-1}$: it is $-x^{-2}$. In algebra, every derivation $D \colon R \to R$ on an integral domain $R$ extends uniquely to a derivation on the fraction field $K$ of $R$. This unique extension is the quotient rule turned into a definition: $D(f/g) = (gD(f) - fD(g))/g^2$. So the quotient rule is not useless. You have to check this is well-defined. It is motivated by the product rule (apply the product rule with $D$ to $(g)(f/g) = f$ as if $D(f/g)$ exists and then solve for $D(f/g)$). For $R = \mathbf R[x]$ and $D = d/dx$, $D(x^n) = nx^{n-1}$ for all integers $n$. $\endgroup$
    – KCd
    May 22, 2021 at 5:17
  • $\begingroup$ @KCd: Thanks for your comment. I am not sure how I accidentally confused it with something else, and certainly I agree with you on the formal derivative of $x^{-1}$... I've edited to remove the error. $\endgroup$
    – user21820
    May 22, 2021 at 5:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.