Question

differentiate ln(secx)....

is it 1/secx • √tanx ?

Answer 1

shoot i closed the question accidentally agh

Answer 2

@hartnn heya :3 any idea?^^

Answer 3

take the derivative? yeah

Answer 4

was my answer correct?

Answer 5

what is the derivative of ln(u) with respect to x ?

Answer 6

1/secx • √tanx ?

Answer 7

or is it just √tanx ?

Answer 8

\[ln(u)\to \frac{u'}{u}\]

u = tan(x)

Answer 9

is there any way to do it without subbing an unknown?

Answer 10

its not that its an unknown, but this is just an application of the chain rule

Answer 11

but the chain rule doesn't need substitution...

Answer 12

f(g) = f'(g) * g'

f = ln(x), g=tan(x)

Answer 13

the chain rule is present in all the rules youve covered. you just havent realized it

Answer 14

i use the chain rule, its like one of the three rules, but i don't substitute anything ><

Answer 15

the derivative of ln(x) is x'/x

what is x' with repect to x tho? dx/dx = 1

Answer 16

you are using a mental shortcut when you 'dont substitute' anything.

Answer 17

isn't it 1/x?

Answer 18

i mean, the formula they give me is basically

dy/dx = d/dx (lnx) = 1/x

Answer 19

only with resect to x

what is x' with respect to time? dx/dt which isnt always going to equal 1

Answer 20

why do you put a prime behind the x?

Answer 21

its a simpler notation to me

Answer 22

what does it mean?

Answer 23

x' is the derivative of x with respect to something.

x' w.r.t x = dx/dx = 1
x' w.r.t. t = dx/dt
x' w.r.t. g = dx/dg
etc...

Answer 24

okay

Answer 25

just as y' is the derivative of y w.r.t something

y' w.r.t x = dy/dx
y' w.r.t t = dy/dt
y' w.r.t y = dy/dy = 1

etc ...

Answer 26

this is how implicit derivatives are worked

Answer 27

okay

Answer 28

so

chain rule: f(g) derives to f'(g) * g'

f = ln(x), g=tanx

f' = 1/tanx, g' = sec^2(x)

Answer 29

d/dx(tanx)=sec^2x

Answer 30

lol, forgot what g was

Answer 31

ln(secx)

f = ln(x) and g = secx

Answer 32

so what would integrate ln(secx) be?

Answer 33

sec tan
------ = tan
sec

Answer 34

so basically, integrate ln(secx) = integrate ln(sec^2(x))

Answer 35

not quite

ln(a^b) = b ln(a)

so the derivative of ln(sec^2) = 2 ln(sec)

Answer 36

that wasnt worded right

Answer 37

the derivative of ln(sec^2) = the derivaitve of2 ln(sec)

Answer 38

how do you integrate sec?

Answer 39

sec was integrated long before integration was established :)

Answer 40

whaaat but my question was
how do you integrate ln(secx)?

Answer 41

wait so its actually 1/2 tan?

Answer 42

multiply by a useful form of 1
\[sec=\frac{sec^2+sectan}{sec+tan}\]

Answer 43

im not sure we are on the same page here.

what question are you asking at the moment?

Answer 44

what is integrate ln(secx)?

Answer 45

integrate ln(sec(x))

as opposed to the derivative of ln(sec(x))

right?

Answer 46

sorry differentitate i mean ><

Answer 47

whew, cuase the integration is ugly

Answer 48

wow i'm confusing us all sorry xD

Answer 49

lets take it in steps

do you agree that we define the chain rule as:

[f(g(x))] = f'(g(x)) times g'(x) ?

Answer 50

okay hang on could you just write the answer here first :3

Answer 51

i already did

sec tan
------ = tan
sec

Answer 52

so differentiate ln(secx) is tanx, to confirm?

Answer 53

correct

Answer 54

therefore differentiate secx is secxtanx?

Answer 55

g' = secx tanx correct

Answer 56

what would differentiate sec^3x be then?

Answer 57

(secx)^3

f = ()^3 and g = secx

apply chain rule

Answer 58

ah

Answer 59

can't i do

sec^3x
=3sec^2x(3secxtanx)?

Answer 60

yes, but without the extra 3 in there

Answer 61

oh?

Answer 62

f' = 3()^2 g' = sec tan

Answer 63

oh yes because when i take the derative i have to take simply secx right?

Answer 64

yes

Answer 65

phew thank you :3

Answer 66

youre welcome :)