Question

use logarithmic differntiation:

Answer 1

to find y given :
\[y=(\tan \theta)\sqrt{2\theta+1)}\]

Answer 2

all you need to do is log both sides

Answer 3

can you show?

Answer 4

log y = logt(theta) + 1/2 log (2theta +1)

Answer 5

im here to learn!

Answer 6

\[ln(y)=\ln[\tan(\theta)*\sqrt{2\theta+1}]\]
\[\ln(y)=\ln(\tan(\theta))+\ln(2\theta+1)^\frac{1}{2}\]
\[\ln(y)=\ln(\tan(\theta))+\frac{1}{2}\ln(2\theta+1)\]
take derivative of both sides
\[\frac{y'}{y}=\frac{\sec^2(\theta)}{\tan(\theta)}+\frac{1}{2}*\frac{2}{2\theta+1}\]

Answer 7

then when you diff log y you get 1/y *dy/dx =rhs

Answer 8

now our objective is to find y' so multiply y on both sides and woolah
hey joe
did you learn something lol

Answer 9

interesting! i have never heard of that technique before, but it makes sense lol

Answer 10

lol
i dont believe you

Answer 11

totally serious! i was like, "logarithmic differentiation......wut"

Answer 12

logarithmic differentiation can be really useful for some functions

Answer 13

but its pretty much what it sounds like.

Answer 14

can someone write it out for me:) i will be eternally grateful

Answer 15

yeah thats what i was going to ask next, is there a particular case where you would look at a problem and go, "this problem SCREAMS for Log diff."

Answer 16

\[y=sinx^{cosx}\]
how would you find y' here then?

Answer 17

take ln of both sides?

Answer 18

this one screams it badly

Answer 19

oh yeah. i would take ln of both sides. I guess ive just never called it by its name lol.

Answer 20

\[\ln(y)=\ln(sinx)^{cosx}\]
\[\ln(y)=cosx*\ln(sinx)\]
\[\frac{y'}{y}=-sinx*\ln(sinx)+cosx*\frac{cosx}{sinx}\]

Answer 21

what would the final anser be for the original problem i posted

Answer 22

it would be what i have above multiplied by y (y=tan(theta)sqrt{2(theta)+1}

Answer 23

lol

Answer 24

wait what you orginal posted of the y=sinx^cosx

Answer 25

now i was talking about your problem not the one i was doing for example

Answer 26

i am confused, i still dont see what the final answer would be

Answer 27

do you now see what i did for your problem?

Answer 28

not*

Answer 29

yea, above, before you did the example problem of y=sinx^cosx right?

Answer 30

yes

Answer 31

i got to y'/y=something
well y' would just be that (something)*y

Answer 32

so, you are saying multiply all of that by the orginal problem

Answer 33

ok above when i was doing your problem i got
y'/y=something right?

Answer 34

right, cause we are trying to find the derivative so, we multiply both sides by y

Answer 35

right!

Answer 36

so we would then have y'=(something)*y right?

Answer 37

yes

Answer 38

help with one more??

Answer 39

make another post this one is getting long k? :)