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DLSBest ResponseYou've already chosen the best response.1
\[\LARGE \frac{1}{y} \times \frac{dy}{dx} =\frac{cosx}{sinx} \times (sinx) + (cosx)(cosec^{2}x) + \frac{sinx}{cosx} \] \[\LARGE \times (cosx)+(sinx)(\sec^{2}x)\]
 one year ago

DLSBest ResponseYou've already chosen the best response.1
last term hidden is sinx/cos then continued below
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
what was that ?!! y= u+v find du/dx and dv/dx using logarithmic differentiation then add...
 one year ago

DLSBest ResponseYou've already chosen the best response.1
I used chain rule for logsinx then UV rule with (cosx)logsinx
 one year ago

DLSBest ResponseYou've already chosen the best response.1
\[\LARGE logy=(cosx)(logsinx)+(sinx)(logcosx)\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
log (A+B) cannot be simplified to log A +log B
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
y= u+v find du/dx and dv/dx using logarithmic differentiation then add... thats the only way...
 one year ago

DLSBest ResponseYou've already chosen the best response.1
\[\LARGE logy=\log(cosx^{cosx} \times sinx + (sinx^{sinx} \times cosx)\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
u =sin x^cos x du/dx =... ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
again ? this is the 1st time, you use log...
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
log u = cos x log sin x now diff. use product rule.
 one year ago

DLSBest ResponseYou've already chosen the best response.1
thats what I did.. I got: cotx cosx
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
did u forget product rule ?
 one year ago

DLSBest ResponseYou've already chosen the best response.1
\[cosxlogsinx= cosx(\frac{1}{sinx}) + logsinx (sinx)\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
1/u du/dx = cos x (1/ sin x) (cos x)  sin x log (sinx)
 one year ago

DLSBest ResponseYou've already chosen the best response.1
I guess I studied too much :P
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
extra cos x come due to chain rule is it ?
 one year ago

DLSBest ResponseYou've already chosen the best response.1
cosxlogsinx =>cosx(1/sinx * cosx)(chain rule ) cosxcotx now product rule whats wrong with it :/
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
because there was no 2nd term in there....thats why i asked u about product rule...
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
ok, just tell me what u get finallu for du/dx .. ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
u get this, right ? 1/u du/dx = cos x (1/ sin x) (cos x)  sin x log (sinx)
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
don't write in latex, takes time unnecessarily...
 one year ago

DLSBest ResponseYou've already chosen the best response.1
yes,im solving last time :o only du/dx
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
yes, only du/dx....dv/dx will be very similar....
 one year ago

DLSBest ResponseYou've already chosen the best response.1
cosx cosec^2 x sinxcotx :///////
 one year ago

DLSBest ResponseYou've already chosen the best response.1
on solving cosxcotx by product rule thats it
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
:( didn't u get this ? 1/u du/dx = cos x (1/ sin x) (cos x)  sin x log (sinx)
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
log u = cos x log sin x this ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
(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 ?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
just simple product rule.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
want to continue ? or should i ?
 one year ago

DLSBest ResponseYou've already chosen the best response.1
logsinx(sinx)+cosx(logsinx) okay?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
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 ? ^
 one year ago

DLSBest ResponseYou've already chosen the best response.1
\[\LARGE logsinx(sinx)+cosx(\frac{1}{sinx} \times cosx)\] RIGHT?
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
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 ?
 one year ago

DLSBest ResponseYou've already chosen the best response.1
\[FINAL:logsinx(sinx)+cosxcot\]
 one year ago

DLSBest ResponseYou've already chosen the best response.1
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
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
du/dx = [sin x^cos x][cos x . cot x  sin x log sin x] in same manner find dv/dx.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
and to get final answer, just add them
 one year ago

DLSBest ResponseYou've already chosen the best response.1
that was one hell of a question _'
 one year ago

DLSBest ResponseYou've already chosen the best response.1
it was new for me..so yeah..next time easy for me too...(hopefully)
 one year ago
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