Dude I don't know what you're trying to get at lol...Can you just show me

 

i wont do it for you, i will do examples though

super simple

f(x) = 4x(2x^2 - 5)^3

what you need to look for is a set of terms that one is the deriv of the other
2x^2's deriv is 4x

but also 2x^2 - 5 's deriv is also 4x

which one makes our substitution more useful though?

lets try both

first
let u=2x^2
then du=4x
we get du(u - 5)^3
is that any better than our original? nope

ok, second
what about let u = 2x^2 -5
then du = 4x
so we have f(x) = du (u)^3

well shit we can do that no problem

see the connection?

 

U = 16-x^2

U' = -2x

U^1/2 du

2/3U^3/2

2/3(16-x^2)^3/2

 

:yay:

nicely done

 

The square root was fuckin me up

 

the square root is what fucks up being able to do it without substitution

have you done u,v sub yet?
that shit gets even more complicated, need to keep track of things and keep it simple

 

theres only two classes left so i doubt we will get into it

 

well its basically the next chapter, but unless he plans on fucking you guys over i wouldnt think he would do it right before the final

get #3 ok?

i would practice the substitution stuff a little more before the test, its a very common type of problem, i would bet at least one of them shows up on the test that was almost exactly like that one, very common with square roots and also ugly looking division
but it simplifies so nicely when you get it, so its just about seeing the trick

 

Yeah, he said he was puttin a substitution problem on the final...I haven't had a chance to go over #3 yet...I'll do it tonight lol

 

http://www.imeem.com/albdeno/music/u...on-cant-count/