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DLS
 4 years ago
y=sinx^cosx+cosx^sinx
dy/dx=?
DLS
 4 years ago
y=sinx^cosx+cosx^sinx dy/dx=?

This Question is Closed

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

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1last term hidden is sinx/cos then continued below

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

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1I used chain rule for logsinx then UV rule with (cosx)logsinx

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1\[\LARGE logy=(cosx)(logsinx)+(sinx)(logcosx)\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1log (A+B) cannot be simplified to log A +log B

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

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1\[\LARGE logy=\log(cosx^{cosx} \times sinx + (sinx^{sinx} \times cosx)\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1u =sin x^cos x du/dx =... ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1again ? this is the 1st time, you use log...

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1log u = cos x log sin x now diff. use product rule.

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1thats what I did.. I got: cotx cosx

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1did u forget product rule ?

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1\[cosxlogsinx= cosx(\frac{1}{sinx}) + logsinx (sinx)\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.11/u du/dx = cos x (1/ sin x) (cos x)  sin x log (sinx)

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1I guess I studied too much :P

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1extra cos x come due to chain rule is it ?

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

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1because there was no 2nd term in there....thats why i asked u about product rule...

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1ok, just tell me what u get finallu for du/dx .. ?

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

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1don't write in latex, takes time unnecessarily...

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1yes,im solving last time :o only du/dx

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1yes, only du/dx....dv/dx will be very similar....

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1cosx cosec^2 x sinxcotx :///////

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1on solving cosxcotx by product rule thats it

hartnn
 4 years ago
Best 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)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1log u = cos x log sin x this ?

hartnn
 4 years ago
Best 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 ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1just simple product rule.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1want to continue ? or should i ?

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1logsinx(sinx)+cosx(logsinx) okay?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1my 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 ? ^

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1\[\LARGE logsinx(sinx)+cosx(\frac{1}{sinx} \times cosx)\] RIGHT?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.11/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
 4 years ago
Best ResponseYou've already chosen the best response.1\[FINAL:logsinx(sinx)+cosxcot\]

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1this 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
 4 years ago
Best ResponseYou've already chosen the best response.1du/dx = [sin x^cos x][cos x . cot x  sin x log sin x] in same manner find dv/dx.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.1and to get final answer, just add them

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1that was one hell of a question _'

DLS
 4 years ago
Best ResponseYou've already chosen the best response.1it was new for me..so yeah..next time easy for me too...(hopefully)
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