DLS
  • DLS
y=sinx^cosx+cosx^sinx dy/dx=?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DLS
  • DLS
My attempt:
DLS
  • DLS
\[\LARGE \frac{1}{y} \times \frac{dy}{dx} =\frac{cosx}{sinx} \times (-sinx) + (cosx)(-cosec^{2}x) + \frac{sinx}{cosx} \] \[\LARGE \times (cosx)+(sinx)(\sec^{2}x)\]
DLS
  • DLS
last term hidden is sinx/cos then continued below

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More answers

hartnn
  • hartnn
what was that ?!! y= u+v find du/dx and dv/dx using logarithmic differentiation then add...
DLS
  • DLS
lol
DLS
  • DLS
I used chain rule for logsinx then UV rule with (cosx)logsinx
DLS
  • DLS
\[\LARGE logy=(cosx)(logsinx)+(sinx)(logcosx)\]
hartnn
  • hartnn
log (A+B) cannot be simplified to log A +log B
DLS
  • DLS
where is log(A+B) :O
DLS
  • DLS
oh
DLS
  • DLS
NEVERMIND
hartnn
  • hartnn
y= u+v find du/dx and dv/dx using logarithmic differentiation then add... thats the only way...
abb0t
  • abb0t
Use chain rule.
DLS
  • DLS
\[\LARGE logy=\log(cosx^{cosx} \times -sinx + (-sinx^{sinx} \times cosx)\]
DLS
  • DLS
?
hartnn
  • hartnn
u =sin x^cos x du/dx =... ?
DLS
  • DLS
again log? O_O
DLS
  • DLS
okay no
hartnn
  • hartnn
again ? this is the 1st time, you use log...
DLS
  • DLS
:/
DLS
  • DLS
cosxlogsinx?
DLS
  • DLS
CONFUSED
hartnn
  • hartnn
log u = cos x log sin x now diff. use product rule.
DLS
  • DLS
thats what I did.. I got: cotx cosx
hartnn
  • hartnn
did u forget product rule ?
DLS
  • DLS
:/ no :///
hartnn
  • hartnn
(uv)' = uv'+u'v
DLS
  • DLS
\[cosxlogsinx= cosx(\frac{1}{sinx}) + logsinx (-sinx)\]
DLS
  • DLS
isnt it :/..
hartnn
  • hartnn
1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)
DLS
  • DLS
I guess I studied too much :P
DLS
  • DLS
thats what I did >.<
hartnn
  • hartnn
extra cos x come due to chain rule is it ?
DLS
  • DLS
cosxlogsinx =>cosx(1/sinx * cosx)(chain rule ) cosxcotx now product rule whats wrong with it :/
hartnn
  • hartnn
because there was no 2nd term in there....thats why i asked u about product rule...
DLS
  • DLS
stil confused :|
hartnn
  • hartnn
ok, just tell me what u get finallu for du/dx .. ?
hartnn
  • hartnn
u get this, right ? 1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)
DLS
  • DLS
let me write
hartnn
  • hartnn
don't write in latex, takes time unnecessarily...
DLS
  • DLS
yes,im solving last time :o only du/dx
DLS
  • DLS
i get
hartnn
  • hartnn
yes, only du/dx....dv/dx will be very similar....
DLS
  • DLS
-cosx cosec^2 x -sinxcotx :///////
DLS
  • DLS
on solving cosxcotx by product rule thats it
hartnn
  • hartnn
:( didn't u get this ? 1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)
DLS
  • DLS
I dont think so loll :/
hartnn
  • hartnn
log u = cos x log sin x this ?
DLS
  • DLS
yes
DLS
  • DLS
cos x (1/sinx * cosx)
DLS
  • DLS
log dissapeared
hartnn
  • hartnn
(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 ?
hartnn
  • hartnn
just simple product rule.
DLS
  • DLS
okay
hartnn
  • hartnn
want to continue ? or should i ?
DLS
  • DLS
i will
DLS
  • DLS
logsinx(-sinx)+cosx(logsinx) okay?
DLS
  • DLS
DUH
hartnn
  • hartnn
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 ? ^
hartnn
  • hartnn
use of chain rule.
DLS
  • DLS
1minute and its over!
DLS
  • DLS
give me
DLS
  • DLS
lol been 30 minutes:/
hartnn
  • hartnn
let me do it...
hartnn
  • hartnn
u just see.
DLS
  • DLS
ivedone
DLS
  • DLS
\[\LARGE logsinx(-sinx)+cosx(\frac{1}{sinx} \times cosx)\] RIGHT?
hartnn
  • hartnn
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 ?
DLS
  • DLS
\[FINAL:logsinx(-sinx)+cosxcot\]
DLS
  • DLS
x
DLS
  • DLS
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
hartnn
  • hartnn
du/dx = [sin x^cos x][cos x . cot x - sin x log sin x] in same manner find dv/dx.
hartnn
  • hartnn
and to get final answer, just add them
DLS
  • DLS
YES! correct!
DLS
  • DLS
that was one hell of a question -_-'
hartnn
  • hartnn
but easy.
DLS
  • DLS
for u!
hartnn
  • hartnn
sum of 2 eaSY Q's
DLS
  • DLS
it was new for me..so yeah..next time easy for me too...(hopefully)

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