Question

I need to to show that, using the quotient rule:

g(x)=
(e(-3x+2sinx)/6)
----------------
(3+2cosx)

has derivative

g'(x)=
- (4(sin^2)x -12sin x +5) (e(-3x+2sinx)/6) ----------------------------------- 6(3+2cosx)^2

Answer 1

\[g(x)=\frac{e ^{\frac{-3x+2\sin x}{6}}}{3+2\cos x}\]

Answer 2

Is that the problem?

Answer 3

yes

Answer 4

\[\frac{u}{v}=\frac{u'v-v'u}{v^2}\]

Answer 5

\[(\frac{u}{v})'\]

Answer 6

The problem is to simplify to get the deritive as expressed. Could you show me the complete working out

Answer 7

Denominator = \[(3+2\cos x)^2\]
Numerator :
\[u'=(\frac{-1}{2}+\frac{1}{3}\cos x)e^{\frac{-3x+2\sin x}{6}}\]
\[v'=-2\sin x\]
Notice that : \[(\frac{-1}{2}+\frac{1}{3}\cos x)(3+2\cos x)=\frac{4\cos^2 x - 9}{6}\]
Factor out \[e^{\frac{-3x+2\sin x}{6}}\]
Thus numerator = \[e^{\frac{-3x+2\sin x}{6}}(\frac{4\cos^2 x-9}{6}+2\sin x)\]
Notice again : \[\cos^2 x +\sin^2 x=1\]
Hence numerator = \[e^{\frac{-3x+2\sin x}{6}}(\frac{-5-4\sin^2 x+12\sin x}{6})\]
.... :)

Answer 8

Thanks for your help.

Answer 9

you're welcome ;)

Answer 10

Hi anhkoavo1210

Answer 11

Hi

Answer 12

Ok i'm here

Answer 13

which part in numerator you want I explain?

Answer 14

Could you show me how the how I get from e(-3x+2sinx)/6) to
- (4(sin^2)x -12sin x +5) (e(-3x+2sinx)/6)

Answer 15

I can do some of it with the quotient rule but get stuff in the simpplification process

Answer 16

get stuck with the simplification process

Answer 17

uh, can you display your numerator you're having

Answer 18

I get to this point
(2cosx+3)(e^(2sinx-3x)/6)(2cosx-3) - (e^(2sinx-3x))(-2sinx)
---------------------
6
Then i get stuck

Answer 19

2cosx+3)(e^(2sinx-3x)/6)(2cosx-3) - (e^(2sinx-3x)/6)(-2sinx)
---------------------
6

Answer 20

that is what it should be - sorry typing is difficult

Answer 21

Ok, I write down again :
\[(\frac{-1}{2}+\frac{1}{3} \cos x)e^{\frac{−3x+2sinx}{6}}(3+2\cos x)+e^{\frac{−3x+2sinx}{6}}2\sin x\]
You got it here?

Answer 22

How did you get -1/2 + 1/3 cosx

Answer 23

You try to see \[(e^{\frac{−3x+2sinx}{6}})'=?\]

Answer 24

Ok!

Answer 25

And how to I get to my derivative using cos^2+sin^2x = 1

Answer 26

for the last part as you mentioned before

Answer 27

Did you got it here?
\[e^{\frac{−3x+2sinx}{6}}(\frac{4\cos^2 x−9}{6}+2\sin x)\]

Answer 28

Can explain how you do this part?

Answer 29

you see\[ (−\frac{1}{2}+\frac{1}{3}\cos x)(3+2cosx)=\frac{4\cos^2x−9}{6}\] ??

Answer 30

you factored?

Answer 31

expanded

Answer 32

no, you see \[\frac{−1}{2}+\frac{1}{3}\cos x=\frac{-3+2\cos x}{6}\]
After I use formula : \[(a-b)(a+b)=a^2-b^2\]

Answer 33

OK

Answer 34

Thanks for you help. I have to sit down and spend time on this. Thanks very much. I am making some progress. Do you min if I as for some help in the future. I am also and English teacher so if you need help from me then I more then happy to help, even conversation where I can correct you English expression even over chat. Let me know. I have to go now.

Answer 35

I meant if you mind if I ask for more help in the next few days as I have an exam soon.

Answer 36

:D ok, thank you. Do you have a nickname yahoo?

Answer 37

Are you on skype?

Answer 38

No i usually use yahoo.

Answer 39

this is my email address on yahoo
hussain_m_ismail@yahoo.co.uk

Answer 40

tell me your name :D

Answer 41

I have to go now so keep in touch. And I will contact you if i need more help, if that is ok

Answer 42

Ok, if I available. But your name is?