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Category Archives: English

Motivation, AND: Calculus in English — the chain rule

One very good reason to start a math blog is that it motivates you to crack the math book slightly more often, simply to have something to blog about. Because if you don’t blog frequently, people won’t keep coming back, and nobody will add you to their blogrolls, and you will die at age 30, alone, cold, wet, and poor in a back alley in Düsseldorf.

So the other major bit of calculus so far that took me inordinately long to parse in text because of the sheer number of symbols boomeranging about was the chain rule. As before, here is my (no doubt amusingly wrong) attempt to translate it into English. Here be a good online symbol-version. Here’s another. Hopefully one of those will make sense to you. (It mildly squicks me to link the second version, as I seem to have dated one of the people on that project who may have created it. This is, like, math-cest. Ewww. This TMI comes to you courtesy of The Blogosphere, Bringing You Irrelevant Personal Revelations Since 2003(r)).

So, ANYWAY! You have zis function, and you want to take its derivative. Only, it’s a little messy (clearly the root of all evil). Lucky for you, it can be written as one function inside another function. For example, consider the function F(x) = (x2 + πx)4. This can be seen as one function — g(x) = x2 + πx inside a second function f(u) = u4.

(Isn’t it interesting how whenever mathematicians want to notate a second function for substitution purposes — in this case replacing the g function with a variable to show how it fits into the f function — they name the big function f(u)? Is this some kind of latent resentment at the world because physics gets all the grant money?)

So here’s what you do to get the derivative of this sucker. You take the derivative of the outside function — f(u) and then replace the u with the inside function, without taking the derivative of the inside function. In the example, f'(u) = 4u3 by the power rule, and that is expanded to 4(x2+πx)3. Then you multiply that result by the derivative of the inside (g, or u) function. Which, of course, is g'(x) = 2x+π. So the final derivative of the whole big function is F'(x) = 4(x2+πx)3(2x+π). Shazam!

Yes, I just put pi in this one so I could play with HTML math symbols.

EDIT: Ok wordpress, get the line breaks right. I do not want to hand-write the HTML, but I will if you cross me.

SECOND EDIT: I’m serious here. Stop removing my paragraph tags. Stop it. Now. NOW.

Math Explained in English: Indefinite Integration by Substitution

This is my first attempt to put a symbol-bound bit of math into English, for ease of my own comprehension and amusement of others. It took me something like an hour to parse out the explanation in the textbook because it relied on a dense bunch of nested functions, variables flying about left and right, and so forth. So I’m rewriting it to cut the comprehension time down to about ten minutes or so.
First, you should see the symbol-bound version. This guy’s web page is incredibly good on the subject.

Here’s my summary.

So, you’ve got this function you need to integrate. Suppose it’s kind of messy, like say [imagine an integration symbol here] 20x(x2+5)2 dx. Well, if you can express that as a function (f) of a second function (g), multiplied by the derivative of that second function (or a constant multiple of that derivative), then you can just substitute the variable “u” for that second function, drop the multiplication by the derivative out of your consciousness (changing dx to du to represent it, and keeping any constant multiple), and integrate the resulting much simpler function.

Sooo… in the example I gave, the g function (“u”) would be x2+5. The f function is then u2 (not, shockingly, the band), because that meets the condition described above (the second function being buried within the first function). The derivative of the g function is 2x, which is a tenth of the 20x we conveniently have sitting around collecting dust. So we can express the new function to be integrated as: [integral symbol] u2 (10)du, which, by the constant multiple rule, becomes 10 [integral symbol] u2 du, for which the integral is easy: 10u3/3 + C. Abracadabra!

Take that, textbook!

Math people: did I screw this up? Time to chime in…
Also, who wants to find me a text-sized gif or jpg integral symbol clipart?