Question

y=sinx^cosx+cosx^sinx
dy/dx=?

Answer 1

My attempt:

Answer 2

\[\LARGE \frac{1}{y} \times \frac{dy}{dx} =\frac{cosx}{sinx} \times (-sinx) + (cosx)(-cosec^{2}x) + \frac{sinx}{cosx} \]
\[\LARGE \times (cosx)+(sinx)(\sec^{2}x)\]

Answer 3

last term hidden is sinx/cos then continued below

Answer 4

what was that ?!!
y= u+v
find du/dx and dv/dx using logarithmic differentiation
then add...

Answer 5

lol

Answer 6

I used chain rule for logsinx then UV rule with (cosx)logsinx

Answer 7

\[\LARGE logy=(cosx)(logsinx)+(sinx)(logcosx)\]

Answer 8

log (A+B) cannot be simplified to log A +log B

Answer 9

where is log(A+B) :O

Answer 10

oh

Answer 11

NEVERMIND

Answer 12

y= u+v
find du/dx and dv/dx using logarithmic differentiation
then add...
thats the only way...

Answer 13

Use chain rule.

Answer 14

\[\LARGE logy=\log(cosx^{cosx} \times -sinx + (-sinx^{sinx} \times cosx)\]

Answer 15

?

Answer 16

u =sin x^cos x
du/dx =... ?

Answer 17

again log? O_O

Answer 18

okay no

Answer 19

again ? this is the 1st time, you use log...

Answer 20

:/

Answer 21

cosxlogsinx?

Answer 22

CONFUSED

Answer 23

log u = cos x log sin x
now diff.
use product rule.

Answer 24

thats what I did..
I got:
cotx cosx

Answer 25

did u forget product rule ?

Answer 26

:/ no :///

Answer 27

(uv)' = uv'+u'v

Answer 28

\[cosxlogsinx= cosx(\frac{1}{sinx}) + logsinx (-sinx)\]

Answer 29

isnt it :/..

Answer 30

1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)

Answer 31

I guess I studied too much :P

Answer 32

thats what I did >.<

Answer 33

extra cos x come due to chain rule
is it ?

Answer 34

cosxlogsinx
=>cosx(1/sinx * cosx)(chain rule )
cosxcotx
now product rule
whats wrong with it :/

Answer 35

because there was no 2nd term in there....thats why i asked u about product rule...

Answer 36

stil confused :|

Answer 37

ok, just tell me what u get finallu for du/dx .. ?

Answer 38

u get this, right ?
1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)

Answer 39

let me write

Answer 40

don't write in latex, takes time unnecessarily...

Answer 41

yes,im solving last time :o
only du/dx

Answer 42

i get

Answer 43

yes, only du/dx....dv/dx will be very similar....

Answer 44

-cosx cosec^2 x -sinxcotx
:///////

Answer 45

on solving
cosxcotx by product rule thats it

Answer 46

:(
didn't u get this ?
1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)

Answer 47

I dont think so loll :/

Answer 48

log u = cos x log sin x
this ?

Answer 49

yes

Answer 50

cos x (1/sinx * cosx)

Answer 51

log dissapeared

Answer 52

(log u)' = (cos x log sin x )'
1/u u' = cos x (log sin x)' + log sin x (cos x)'
did u get this step ?

Answer 53

just simple product rule.

Answer 54

okay

Answer 55

want to continue ?
or should i ?

Answer 56

i will

Answer 57

logsinx(-sinx)+cosx(logsinx) okay?

Answer 58

DUH

Answer 59

my next step
1/u u' = cos x (log sin x)' + log sin x (cos x)'
1/u u' = cos x (1/ sin x)(sin x)' + log sin x (-sin x)
get that ? ^

Answer 60

use of chain rule.

Answer 61

1minute
and its over!

Answer 62

give me

Answer 63

lol been 30 minutes:/

Answer 64

let me do it...

Answer 65

u just see.

Answer 66

ivedone

Answer 67

\[\LARGE logsinx(-sinx)+cosx(\frac{1}{sinx} \times cosx)\]
RIGHT?

Answer 68

1/u u' = cos x (log sin x)' + log sin x (cos x)'
1/u u' = cos x (1/ sin x)(sin x)' + log sin x (-sin x)
1/u u' = cos x (1/ sin x)(cos x) - sin x log sin x

du/dx = u [cos x . cot x - sin x log sin x]
finally
du/dx = [sin x^cos x][cos x . cot x - sin x log sin x]
any doubts in any step ?

Answer 69

\[FINAL:logsinx(-sinx)+cosxcot\]

Answer 70

x

Answer 71

this is what u get out of cosx log sinx right? ^
and im doing it myself...ill check from urs,this is very confusingnow
just tell me that^
though i checked with WF

Answer 72

du/dx = [sin x^cos x][cos x . cot x - sin x log sin x]
in same manner find dv/dx.

Answer 73

and to get final answer, just add them

Answer 74

YES! correct!

Answer 75

that was one hell of a question -_-'

Answer 76

but easy.

Answer 77

for u!

Answer 78

sum of 2 eaSY Q's

Answer 79

it was new for me..so yeah..next time easy for me too...(hopefully)