Question

find derivative of log

Answer 1

log what?

Answer 2

log x?

Answer 3

\[y=\ln(\sqrt{\sin \theta \cos \theta}/1+2\ln \theta)\]

Answer 4

ahhh, that's more like it

Answer 5

looks annoying...

Answer 6

well, use the chain rule first, so derivative of ln(all that stuff)=

[1/(all that stuff) ][derivative of all that stuff]

Answer 7

Use wolfram:
http://www.wolframalpha.com/input/?i=derivative%3A+ln%28Sqrt%28sin%CE%B8cos%CE%B8%29%2F%281%2B2ln%CE%B8%29%29

Answer 8

so you get something like:

[1+2ln(theta)/sqrt(sin(theta)cos(theta)][more work than i'm willing to put in right now]

Answer 9

will theys how me how to do it

Answer 10

true story with the wolfram! yes, they have a show me how button thingy and you click it it will show you step by step how to do it

Answer 11

they didnt for this one

Answer 12

i feel like i done this one b4

Answer 13

i doubt you would forget it

Answer 14

Is the context implicit differentiation, because that might be faster.

Answer 15

no, natural log derivatives

Answer 16

i would do ln(a/b)=lna-lnb first

Answer 17

i need help, i did it but it looks nothing ,like what wolfram is showing

Answer 18

myin is onto it, break it up with rules of logs

Answer 19

after using ln(a/b) property, use ln(a)^r = rln(a) to remove sqrt

Answer 20

\[y=\ln(sinxcosx)^\frac{1}{2}-\ln(1+2lnx)\]
bring that 1/2 down

Answer 21

then split the sin*cos with the log property of ln(a*b)=lna+lnb

Answer 22

\[y=\frac{1}{2}\ln|sinx|+\frac{1}{2}\ln|cosx|-\ln|1+2lnx|\]

Answer 23

you can bring that one half down and also then you will have to take the deriviative of ln(u) where u is the argument. THis will result in 1/u*d (u)/dx

Answer 24

y'=1/2*cosx/sinx+1/2*(-sinx)/cosx-(2/x)/(1+2lnx)

Answer 25

is that the final answer

Answer 26

yeah looks like it , although mine is in a different form

Answer 27

so what is the correct answer?

Answer 28

what do you mean?
i thought you wanted y'
you can always rewrite y'
make it look pretty, but i gave you y'

Answer 29

could you wriit it using the equation from cause its kinda hard to read it that way you wriote it

Answer 30

if you have y=ln|f(x)|
then y'=f'/f

Answer 31

you try look at the y i wrote above from your y
and use wherever you have ln|f| to find derivative just use f'/f

Answer 32

you can do it
i believe in you

Answer 33

whats the derivative of ln|sinx|?

Answer 34

you still dont get this problem metalcruncher?

Answer 35

no, and its math cruncher. is the derivative 1/sinx*cosx

Answer 36

ok whats the derivative of ln|cosx|?

Answer 37

1/cosx*-sinx

Answer 38

whats the derivative of ln|1+2lnx|?

Answer 39

thats is getting hard for me?

Answer 40

1/1+2lnx*2/x

Answer 41

1/(1+2lnx)*2/x
is right!
gj
you are done

Answer 42

how about that square root

Answer 43

remember we brung the 1/2 down

Answer 44

don't forget about you constant multiples when you put all of this together

Answer 45

right, but shouldnt there still be a -1/2 exopent on that term then

Answer 46

you should have the constant multiples 1/2, 1/2,-1 respectively
for each of the derivative you found :
make sure you multiply the first by 1/2
the second by 1/2
and third by -1
remember (cf)'=cf'

Answer 47

okay let me show you what i came up with

Answer 48

\[1/2*(\cos \theta *(\cos \theta+\sin \theta)*(-\sin \theta))/(\sin \theta \cos \theta))-((2/\theta)/(1+2\ln \theta))\]

Answer 49

and let me show you what i think you said your y is
so that is no misunderstanding
\[\ln(\frac{\sqrt{sinxcosx}}{1+2lnx})\]
is that right?

Answer 50

and know something is wrong with your y'

Answer 51

yeah taht is the correct problem

Answer 52

ok what did you so with derivative of 1/2*ln|sinx| and the derivative of 1/2*ln|cosx|
what is that extra stuff
and you should have three terrms not two

Answer 53

basically i alsoe multiplied by sides by y

Answer 54

why?
derivative of y is just y'

Answer 55

okay, so do you think i would be right i i just took out the term after the subtraction sign

Answer 56

ok looking at the first page
do you have any problem with the algebra i did?

Answer 57

on the second page that should be cot(2x) not tan(2x)

Answer 58

since cos/sin=cot lol

Answer 59

so anyways any problem with the algebra on page 1?

Answer 60

so it should be cot2x

Answer 61

thats right do you know why?

Answer 62

because we have \[\frac{\cos(2x)}{\sin(2x)}\] the step before right?

Answer 63

now the first y' that i wrote is fine
it is still y'
i just tried to rewrite to make the answer look better

Answer 64

that's one long thread, gj

Answer 65

lol yes it is

Answer 66

so true! i can't believe we are still posting to it. lol

Answer 67

mathcruncher do you have any questions about my attachment?

Answer 68

before you say something about my attachment
i already made a correction to it
y'=cot(2x)-2/(x+2xlnx)

Answer 69

medal party!

Answer 70

i spent so long on this with him
i think he lefted me
thats mean

Answer 71

That's was one bored teacher who assign such problem

Answer 72

its not that bad of a problem

Answer 73

Too long and tedious

Answer 74

using some log properties we made it look tons nicer

Answer 75

i didnt leave

Answer 76

ok thanks mathcruncher lol

Answer 77

sorry i was always assume the worse

Answer 78

i silt water on my lap top so i had to go get a napkin to wipe it off

Answer 79

smoke coming out of it?

Answer 80

so, it is okay to multiply that term on the right hand side by x even though we dont do it tot he term on the left

Answer 81

i multiplied by x/x not x

Answer 82

thats what i meant

Answer 83

yes x/x=1 right?
1*fraction=fraction

Answer 84

this thread is so long it takes forever for me to get to the bottom to post something new lol

Answer 85

now, do is it that we get cos2x/sin2x fromsin^2x-cos^2x /2coxsinx

Answer 86

how is that i mean?

Answer 87

how is that i mean?

Answer 88

how is that i mean?

Answer 89

\[\frac{\cos^2x-\sin^2x}{2cosxsinx}\]
is what you mean

\[\cos(2x)=\cos(x+x)=cosx*cosx-sinx*sinx=\cos^2x-\sin^2x\]
\[\sin(2x)=\sin(x+x)=sinx*cosx+sinx*cosx=2sinxcosx\]

Answer 90

oh okay

Answer 91

so the final answer should be cot2x rgiht instead f tan2x

Answer 92

\[y'=\cot(2x)-\frac{2}{x+2xlnx}\]
its not obvious to you why its cot and not tan?

Answer 93

yeah i got that

Answer 94

what can i say you are a star, your awesome thanks for all your help

Answer 95

np
sorry i thought you left me

Answer 96

got a question how does: 1/2ln(sinxcosx) become 1/2[lnsinx+lncosx]