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Thursday, November 6, 2008

Derivation Rules Continued

Okay guys, its me, Justus, the last person on the scribe list, beginning to do this here scribe. Bear with me, as it's gonna be a toughie (since there's no slides :[ )

Alrighty, here we go.

So to begin we let's review some stuff we have done previously with derivatives.

Remember the power function rule? f'(x) = nXn-1?

Lets apply this to some basic functions.

f(x) = x-3 f'(x) = -3/x4
f(x) = x-2 f'(x) = -2/x3
f(x) = x-1 f'(x) = -1/x2
f(x) = x1 f'(x) = 1·x0 = 1
f(x) = x2 f'(x) = 2x
f(x) = x3 f'(x) = 3x2

The fancy way to state the rule (nXn-1) is

d/dx [cf(x)] = cd/dx f(x) where c is a constant. Mostly, this is the exact same, meaning the derivative of f(x) with respect to x.

Now I'm sure you can see a pattern developing, and being able to recognize these patterns will help a bundle in the future (I'm sure of it.)However, what about, if the original function, is slightly more complex then say, f(x) = x2?

How about...

f(x) = (x2 + 1)(x3 - 2)?

I'm sure your wondering how to deal with the above, and to be frank, I'm gonna tell you to the best of my knowledge.

Basically, we'd use the
Product Rule

This rules says,

d/dx[f(x)g(x)] = f'(x)g(x)+ f(x)g'(x). Now the only way I can really show what I mean is to apply it to the above function.

f(x) = (x2 + 1)(x3 - 2)
2x(x3-2)+ 3x2(x2 + 1)
2x4 - 4x+ 3x4 + 3x2
5x4 + 3x2 - 4x

See? Its really not all that hard. Basically, because it was already in a "factored" binomial form, all you really needed to do was distribute, the values of the derivatives of each binomial.

Moving onward to the opposite of the Product Rule, we have the Quotient rule. Never saw that coming huh?

Basically, this is the quotient rule;

d/dx[f(x)/g(x)] = [f'(x)g(x)- g'(x)f(x)]/[g'(x)]2

We also started going into the Chain rule but ran out of time.

Because of this, I shall link ye, to some resources which shall elaborate. The quotient rule in more detail The product Rule in more detail Differentiation rules in general. A little (or rather largeish) reference table of differentiation rules.

most of this stuff can be found on pg 205 of the text also.

One final thing we found out before I call this scribe completed.

We discovered that the derivative of ex IS ex and that if you have f(x) = bx f'(x) = ln b. Cool huh?

Alrighty, I think thats everything, homework for tonight issss...

1,,7,9,13,15,19,21 Section 4.2

Next scribe shall be, benchman, because I dont think I've ever chosen him to be scribe before. :]

Myself out. [funny fact, I ended this blog EXACTLY when the song I was listening to ended lol.]


Rence said...

Nice little, F&F there.
Got some Fun facts.

dkuropatwa said...

Cool way to end a scribe post; what song was it? Can I find it on

If you've already learned how to find the derivative of ln(x) then you may appreciate this elegant little proof of how to find the derivative of e^x.

(BTW, (d/dx)(ln(x)) = 1/x, but you knew that, didn't you?)

This result is really quite amazing. I wonder how you all feel about it? In the past most students have said: meh. But after we explore a little about where this comes from and what the implications of it are ... well, most of them still say: meh. I'll keep trying. See you all a week from Monday!

Mr. K.

Dr. Eviatar said...

We haven't done the derivative of ln yet, but we will on Monday. Cute proof.

I believe I waxed somewhat lyrical about the mysteries of the universe with regards to e ;-).

Nice to see you back, Mr. K!