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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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    It'll be a hard sell anywhere. It's fairly opinionated and broad, as the question currently stands. – 2017-01-16
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    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. – 2017-01-16
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    This sounds like a question suitable (if slightly reformulated) for [Mathematics Educators Stack Exchange](http://matheducators.stackexchange.com/) – 2017-01-16
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    Your choice of username seems to describe your views on the subject rather well, *yourlazyphysicist*... – 2017-01-16
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    My apologies. I have deleted the does anyone agree comment. – 2017-01-16
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    I agree, I always thought the quotient rule for differentiation was an uninteresting special case of the product rule and chain ruling $(.)^{-1}$ – 2017-01-16

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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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    +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. – 2017-01-16
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    Oh and how faces light up when you show a high schooler that the Pythagorean theorem provides most trig formulas – 2017-01-18
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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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    Even better, you can derive them with the less formidable rotation matrix. – 2017-01-18