1. ## Calculus help

Hey guys,iv just started doing calculus and im struggling to simplify these 2 equations any further,can any1 help me?

1. [ 2e^-w multiplied by(1+e^-3w) -2e^-w multiplied by (3e^-3) ] divided by [(1+e^-3)^2]

2. [ 2cos(2Q)(inv3Q) + (3sin2q /3q) ]

thanks guys

2. Originally Posted by Ritchie187
2. [ 2cos(2Q)(inv3Q) + (3sin2q /3q) ]
First off the variables Q and q are technically different. Are they supposed to be the same?

And what function is "inv?"

-Dan

3. Originally Posted by Ritchie187
1. [ 2e^-w multiplied by(1+e^-3w) -2e^-w multiplied by (3e^-3) ] divided by [(1+e^-3)^2]
Unless the e^{-3} terms are supposed to be e^{-3w} all I can think to do is factor a e^{-w} out of the numerator., then add the constant terms in the numerator.

-Dan

4. sory, the q's are suposed to be the same and Inv was suposed to be In for inverse. thanks

5. Originally Posted by Ritchie187
sory, the q's are suposed to be the same and Inv was suposed to be In for inverse. thanks
To be clear: Inv{3Q} = 1/(3Q)?

-Dan

6. the only bit i realy need to simplify in question 2 is the last bit i think, simplify

(3sin2q) divided by 3q. i think the answer should either be, (sin2q)/q or just sin2? what do you think?

7. Originally Posted by Ritchie187
the only bit i realy need to simplify in question 2 is the last bit i think, simplify

(3sin2q) divided by 3q. i think the answer should either be, (sin2q)/q or just sin2? what do you think?
First you should be writing it as
[3sin(2q)]/(3q)

Now, it is NOT true that sin(2q) = 2sin(q) or any other such manipulation: the sine function is NOT linear!!

The best you can do is:
[3sin(2q)]/(3q) = [sin(2q)]/q

-Dan