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I learned differential calculus and its rules (quocient, chain, etc) and I got curious about one thing: What do we lose by not using these rules when differentiating? Obviously I've noted some utility on these rules: When they are employed, differentiating becomes a lot easier - I believe it would be impossible (at least for me) to differentiate without these rules.

But I've used two expressions that I think confine my view to the small amount of knowledge I possess: I said that it is easier and that it would be impossible to differentiate for me. Perhaps that's not the case for someone else. I have two questions:

  • Is my reasoning correct?

  • What would be the challenge (presuming the reasoning I presented is correct) on differentiating without these rules?

beep-boop
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Red Banana
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  • It wouldn't be impossible, since you could just use the limit to get the derivative? More tedious, sure. But still possible. – Rivasa Jul 13 '14 at 19:05
  • Or you can just write a program to do it, ie to generate a list of improving approximations, eg: http://math.stackexchange.com/questions/858901/rate-of-change-question-planes-kites-and-the-such/859208#859208 –  Jul 13 '14 at 19:17
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    Just a remark. There are situations where the rules are practicized while the underlying thought of finding limits has disappeared. Think of the derivatives of polynomials in a ring. – drhab Jul 13 '14 at 19:27
  • A perhaps useful analogy is programming in Assembler, versus using a higher-level language, and a package of useful subroutines. – André Nicolas Jul 13 '14 at 20:50
  • The thing is as long as you have a probable candidate for what you think derivative is you can always prove it is infact is the derivate soley on the definition of derivate (epsilon, delta definition). But when it doesn't become clear what it possible convergence of this limit is from definition or have any prior reasonable guess what limit may be, this can close to impossible to see. On top of this many theories can be derived by use of these properties more than just derivatives. – Kamster Jul 13 '14 at 20:51
  • "Is my reasoning correct?" I don't see any reasoning here. What reasoning are we supposed to be checking for correctness? If it's impossible for you to differentiate otherwise, you don't understand calculus. – symplectomorphic Jul 13 '14 at 20:56

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Well you would basically be reproving the rules for differentation all the time, which does not seem to be very sensible. The reason why one has these rules is that one uses them all the time; but of course one could differentiate without them. I don't think that there is anyone who would find differentation without the rules easier. The more interesting question is why would anyone do this?

  • I agree; the product and quotient rules proven from the definition of differentiation, so you could just use the proof of it in lieu of just immediately using it. – Hayden Jul 13 '14 at 19:17
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    In my view one of the characteristics of 'good mathematics' is that the wheel is not reinvented. – drhab Jul 13 '14 at 19:32
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Strictly speaking, by the end of a really good first semester calculus class, you should be able to derive all the rules of differentiation except for the general power rule. Certain derivations of the exponential and/or logarithm derivatives look circular, but they are not. For example, one might define $\exp$ so that the derivative of $\exp$ is $\exp$, and then prove that there exists a unique $e>0$ so that $\exp(x)=e^x$. This is not circular, it's just backwards from the way we defined everything else.

Doing this every time gets to be extremely repetitive, each problem being basically a series of derivations of the basic rules over and over again.

Incidentally, here's why the general power rule is a mess. For positive integer powers it's not hard to prove the power rule. You can apply the binomial theorem to $(x+h)^n$ and everything works out. Computing the derivative of $x^{-1}$ and using the chain rule gets you negative integer powers. Then you can handle rational powers using a trick with implicit differentiation, since $y=x^{m/n} \Leftrightarrow y^n=x^m$.

But then for irrational powers you have trouble. The cleanest way to do it that I know of is to show that a continuous function is uniquely defined by its values on the rationals and then prove that $x^y \equiv \exp(y \ln(x))$ agrees with the usual definition when $y$ is rational. Then you prove the power rule with the exponential, logarithm, and chain rules. This is typically far too much work for its pedagogical utility, so they typically do not do it.

Ian
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  • Good point you made in your first sentence. There are rules that are more primitive and these can be used to derive the others, right? – Red Banana Jul 16 '14 at 01:00
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    @IgäriaMnagarka For certain definitions of the basic functions, all the rules can be derived directly from the definition. For example, if $\ln(x) \equiv \int_1^x \frac{1}{t} dt$, $\exp$ is defined as the inverse of $\ln$, and the trig functions are defined geometrically, then all the basic rules of differentiation can be proven from the definition of the derivative, with some help from the squeeze theorem (for trig), binomial theorem (for the power rule), and the fundamental theorem of calculus (for the logarithm). – Ian Jul 16 '14 at 15:09
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Well, from a purely numerical perspective the derivative would still exist and be "computable" in some sense (certainly approximable). I suppose one could carry out involved algebraic calculations to get the symbolic/algebraic expression for the derivative but what's the point? The point of these rules is that the calculations don't actually end up changing for say $d/dx f(g(x))$ if $f = \sin(x)$ or $x^2$ or $e^x$ or $\sin(e^{\cos(x)})$ (likewise whatever we put in for $g$).

jxnh
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