Derivative Rag

Wanna learn calculus the fun way? Then check this out! Written and performed by Dr. Eric McDowell, chair of the Math Department at Berry College.

• Category

  Education

[T3n50r]

Now, that's a awesome teacher. Wish i had one of those when i went to school. I might have actually learned something. He looks like the kinda dude that try to find solutions on how to help a student that really needs it. ^___^

[David Spencer]

2:28 Oops! Amazing video though.

[Conrado Bojórquez]

Nice song, I have a derivative test tomorrow, I'll sing the derivative rag.

[chrisandashleyadkins]

This was the same wonderful professor who was known for jumping on the desk in class. One of the all time great Mathematics Profs!!

[MikelZseo]

1:07 for a massive troll face on the girl.

[Seth Hilliard]

You dont get it. Im saying you learn new tricks for integration (integration is part of calculus by the way). But, for the most part, its just that, learning a a few tricks to remember. Difficulty is relative. To me, integration is easy, and learning that was just as easy (or difficult) as learning differentiation. Anyway, all im sayin is people learn things differently, and i guess it was just harder for you than it was for me. Stop bitchin, stop actin like you know everything.

[EudoxusOfCnidus]

Correction of prev. post: Obviously one uses *algebraic* geometry for the algebraic extensions :-)

[EudoxusOfCnidus]

Integration is definitely more difficult than differentiation. (One could argue that division is more difficult than multiplication, but the difference is so small that I prefer not to.) The problem with integration is that your "tricks" don't really work in general. If you don't believe me, please try integrating exp(x^2) using the ordinary methods from calculus.

[Seth Hilliard]

Thats like saying "learn how to divide and multiplication will be easy." Its no different in difficulty. All you have to do is remember some tricks and bam, you know calculus.

[cllewis1]

jazz hands!

[EudoxusOfCnidus]

cont.

What is the integral of exp(x^2)(2x^3-2x^2-1)/(x-1)^2?

Despite looking more complicated, one can show that it is exp(x^2)x/(x-1). This is done by considering towers of differential field extensions. As long as the extensions are transcendental, the algorithm requires relatively little prerequisites, but for algebraic extensions you need to know differential geometry. Further pursuing integration and differential equations leads to differential Galois theory.