Question

Differentiate the function:
G(u)= ln sqrt((3u+6)/(3u-6))

The answer is -2/(u^2 -4) but I am not sure how to get to that answer. Steps with explanations would be appreciated!

Answer 1

\[G(u) =\sqrt{\frac{3u+6}{3u-6}}\]

Answer 2

you can either use the product rule or the quotient rule to solve this problem...

Answer 3

\[G(u)=\ln \sqrt{\frac{ 3u+6 }{ 3u-6 }}\] is the equation. I am solving for G'(u)

Answer 4

oh ok

Answer 5

ok i will move with u step by step

Answer 6

STEP ONE SIMPLIFY THE LOG

Answer 7

\[\ln(3u+6)^{1/2}-\ln(3u-6)^{1/2}\]
Thats as far as I got

Answer 8

simplify it even further what happens to the power?

Answer 9

oh, it goes in front, duh

Answer 10

I forget what to do from here though. ln is not a natural number, so i cant use chain rule. \[\frac{1}{2}\ln(3+6)-\frac{1}{2}\ln(3-6)\]

Answer 11

ok what is the diff of 1/2ln(x)?

Answer 12

well u can use the chaing rule

Answer 13

d(ln(x)/dx=1/x

Answer 14

ah, I missed that. I must have not written it in my notes.

Answer 15

so????

Answer 16

so i use chain rule for both the 1/2ln(3u+6) and 1/2ln(3u-6) separately? Im stuck trying to figure out how to deal with the ln

Answer 17

is it just (1/2)(1/3u+6) - (1/2)(1/3u-6)?

Answer 18

ok i will show u

Answer 19

can u see the 3 next to the u

Answer 20

or is it \[[\frac{ 1 }{ 2 }(\frac{ 1 }{ 3u+6 })+\ln(3u+6)] -[\frac{ 1 }{ 2 }(\frac{ 1 }{ 3u-6 })+\ln(3u-6)]\]

Answer 21

yes

Answer 22

look carefully

Answer 23

i will solve it for u

Answer 24

you can factor it out, but is it unnecessary?

Answer 25

the last bit you did was right (1/2)(1/3u+6) - (1/2)(1/3u-6)

Answer 26

now the right thing is (1/2)(3/3u+6) - (3/2)(1/3u-6)

Answer 27

can u see the 3s

Answer 28

yes, where did you get them from? did you just put in (1/3) and (3/1) to keep it balanced?

Answer 29

from my pocket hehehe, well by using the chain rule

Answer 30

look let x = 3u+6, dx =3du agree

Answer 31

ln(u) -diff-> 1/u

Answer 32

right, i get ln(u) => 1/u, but what about the dx=3du?

Answer 33

oh, you are just saying that dx where x=3u+6 is equal to 3

Answer 34

and that is where the 3 comes from because of chain rule!

Answer 35

the cahin rule\[\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}\]

Answer 36

have u seen this rule

Answer 37

not written like that, can you express it in F(x) and G(x), or u and v?

Answer 38

i have in my notes [f(x)/g(x)]' = (f'(x)g(x)-g'(x)f(x))/g(x)^2

Answer 39

or [u/v]' = (u'v-v'u)/v^2

Answer 40

(fog)'=g'(x)f'(g(x))

Answer 41

how would u differentiate (1+2x)^2

Answer 42

wait, i think i got it

Answer 43

by the way we are done with the solution, if you simplify the fraction you would get the required answer

Answer 44

so, 1/2 ln(3u+6) =>
g(x)=3u+6
f(x)=1/2ln(g(x))

Answer 45

omg, yes. thank you so much!

Answer 46

you're more than welcome, this was good practicing .... all the best