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

Mathematics
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My attempt:
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\[\LARGE \frac{1}{y} \times \frac{dy}{dx} =\frac{cosx}{sinx} \times (-sinx) + (cosx)(-cosec^{2}x) + \frac{sinx}{cosx} \] \[\LARGE \times (cosx)+(sinx)(\sec^{2}x)\]
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last term hidden is sinx/cos then continued below

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Other answers:

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

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