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?
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15$\begingroup$ 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-booMay 21, 2021 at 17:09
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1$\begingroup$ How do you think the quotient rule is derived? $\endgroup$– Adam KarlsonMay 21, 2021 at 17:12
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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. $\endgroup$– obscuransMay 21, 2021 at 17:17
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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. $\endgroup$– jjagmathMay 21, 2021 at 17:37
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1$\begingroup$ As said, the quotient rule is not necessary essentially because we can prove it in place whenever we need it. But in this same vein, any result that follows from some set of axioms is not technically necessary if you just assume those axioms. $\endgroup$– SluggerJul 19, 2021 at 12:29
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5 Answers
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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.
answered May 21, 2021 at 17:17
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6$\begingroup$ IOW it's not necessary, just a shortcut result $\endgroup$ May 21, 2021 at 17:18
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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?
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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.
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$\begingroup$ Interesting anecdote (can you tell us who it was). I am sympathetic to the quotient rule because its form suggests some pleasing symmetries and asymmetries. On a related vein, I never memorized the formula for finding the roots of a quadratic, because completing the square is informative, quick and reliable. The students who I tutored at an exercise class at the LSE in the late 80s thought I was mad, but I always got the right answer! $\endgroup$ Jul 11, 2022 at 20:30
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$\begingroup$ @RobArthan I prefer to keep the identity private since that person is still alive. $\endgroup$– KCdJul 12, 2022 at 0:34
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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 $v_k$.
answered May 22, 2021 at 5:58
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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.
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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 $n$. $\endgroup$– KCdMay 22, 2021 at 5:17
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$\begingroup$ @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. $\endgroup$ May 22, 2021 at 5:24