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I think sometimes curriculum contains to many formulae. E.g in calculus why is there a need for the quotient rule when there is the product rule

Any examples for undergrad too anyone can think of?

I think there are more examples which demonstrate taking away some of the profoundness of the result that I have thought about in the past, but do not spring to mind right now

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closed as primarily opinion-based by Adam Hughes, Will R, Winther, JMoravitz, Henrik Jan 16 '17 at 22:18

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ It'll be a hard sell anywhere. It's fairly opinionated and broad, as the question currently stands. $\endgroup$ – pjs36 Jan 16 '17 at 21:36
  • $\begingroup$ M.SE is not a discussion board. As such, questions which include phrases such as "Does anyone agree?" are not suitable for at least two reasons: 1) the question is presumably not being asked rhetorically and answers are likely to be primarily opinion based, and 2) the question is more suited to a proper discussion forum rather than a pure Q&A format. You could delete "Does anyone agree?" from the question, but the rest of the question is off topic and probably too broad anyway. On these grounds, I am voting to close. $\endgroup$ – Will R Jan 16 '17 at 21:53
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    $\begingroup$ This sounds like a question suitable (if slightly reformulated) for Mathematics Educators Stack Exchange $\endgroup$ – Winther Jan 16 '17 at 21:59
  • $\begingroup$ Your choice of username seems to describe your views on the subject rather well, yourlazyphysicist... $\endgroup$ – JMoravitz Jan 16 '17 at 22:06
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    $\begingroup$ I agree, I always thought the quotient rule for differentiation was an uninteresting special case of the product rule and chain ruling $(.)^{-1}$ $\endgroup$ – mathreadler Jan 16 '17 at 23:09
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In my high school, the Pythagorean theorem and the "distance formula" to find the distance between points in $\mathbb{R}^2$ were presented as disjoint concepts, both to be rote-memorized. Ridiculous.

More examples:

  • Can't remember the quadratic formula? No problem. Just start with $ax^2 + bx + c = 0$, divide by $a$, and solve for $x$ by completing the square. The technique of "completing the square" is taught in high school algebra, so this shouldn't be a problem. Instead, they invent mnemonic devices for rote-memorization, and students end up thinking this is "the only way".

  • Trig identities. A lot of them can be derived from manipulating $\sin^2 + \cos^2 = 1$ (itself derivable from the Pythagorean theorem) together with $\displaystyle \tan(x) = \frac{\sin(x)}{\cos(x)}$. There's no need to independently memorize $\tan^2(x) + 1 = \sec^2(x)$, e.g.

  • The "formula" for carrying out integration by parts. If one needs this, it can be re-derived from knowledge of the product rule. That is, suppose we have a product of functions $u(x)v(x)$, taking a derivative yields $\Big( u(x) v(x) \Big)' = u'(x)v(x) + u(x)v'(x)$. Rearranging gives $u(x)v'(x) = \Big( u(x) v(x) \Big)' - u'(x)v(x)$, and integrating both sides gives you want you want.

  • Absolutely no need to memorize the formula for inverse trig function derivatives. For example, suppose we want to know the derivative of $f(x) = \arcsin(x)$. Recall that we have $\sin(f(x)) = x$. An application of the chain rule gives $\cos(f(x))f'(x) = 1$, so we have $\displaystyle f'(x) = \frac{1}{\cos(f(x))}$. Drawing a right triangle and figuring out the sides with the Pythagorean theorem will show $\cos(f(x)) = \sqrt{1-x^2}$.

Unfortunately, math education these days, at least in lower-level courses, is primarily memorization-oriented instead of understanding-oriented. It's incredibly inefficient; it makes people hate math, and it makes students work unnecessarily hard to be successful in their courses.

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  • $\begingroup$ +1 for "Pythagorean theorem and the "distance formula" were presented as disjoint concepts". I went through this very same experience in my high schooling days, it blew my mind when I figured that one was implicitly used in the other. $\endgroup$ – Perturbative Jan 16 '17 at 22:28
  • $\begingroup$ Oh and how faces light up when you show a high schooler that the Pythagorean theorem provides most trig formulas $\endgroup$ – Andres Mejia Jan 18 '17 at 18:48
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Trigonometry is a good example. You can derive all the high school trig formulae from the Euler identity:

$$\exp(i x) = \cos(x) + i \sin(x)$$

E.g.

$$1 = \exp(ix)\times\exp(-i x) = \cos^2(x) + \sin^2(x)$$

$$ \cos(2x) + i \sin(2x) = \exp(2 i x) =\left(\exp(ix)\right)^2 = \cos^2(x)-\sin^2(x) + 2i\sin(x)\cos(x)$$

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  • $\begingroup$ Even better, you can derive them with the less formidable rotation matrix. $\endgroup$ – Logan Luther Jan 18 '17 at 19:54

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