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y=sinx^cosx+cosx^sinx
dy/dx=?
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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
hartnn
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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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\[\LARGE logy=(cosx)(logsinx)+(sinx)(logcosx)\]
hartnn
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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
hartnn
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y= u+v
find du/dx and dv/dx using logarithmic differentiation
then add...
thats the only way...
abb0t
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Use chain rule.
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\[\LARGE logy=\log(cosx^{cosx} \times -sinx + (-sinx^{sinx} \times cosx)\]
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?
hartnn
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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
hartnn
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again ? this is the 1st time, you use log...
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:/
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cosxlogsinx?
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CONFUSED
hartnn
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log u = cos x log sin x
now diff.
use product rule.
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thats what I did..
I got:
cotx cosx
hartnn
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did u forget product rule ?
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:/ no :///
hartnn
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(uv)' = uv'+u'v
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\[cosxlogsinx= cosx(\frac{1}{sinx}) + logsinx (-sinx)\]
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isnt it :/..
hartnn
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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 >.<
hartnn
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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 :/
hartnn
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because there was no 2nd term in there....thats why i asked u about product rule...
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stil confused :|
hartnn
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ok, just tell me what u get finallu for du/dx .. ?
hartnn
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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
hartnn
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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
hartnn
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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
hartnn
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:(
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 :/
hartnn
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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
hartnn
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(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
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just simple product rule.
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okay
hartnn
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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
hartnn
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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
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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:/
hartnn
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let me do it...
hartnn
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u just see.
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ivedone
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\[\LARGE logsinx(-sinx)+cosx(\frac{1}{sinx} \times cosx)\]
RIGHT?
hartnn
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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
hartnn
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du/dx = [sin x^cos x][cos x . cot x - sin x log sin x]
in same manner find dv/dx.
hartnn
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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 -_-'
hartnn
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but easy.
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for u!
hartnn
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sum of 2 eaSY Q's
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it was new for me..so yeah..next time easy for me too...(hopefully)