IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - very simple differentiation qns but cant solve

rakesh61

if y = sin x log cos x

then dy/dx equals to

Ans (cotx log cos x + tan x log sin x ) / ((log cos x )^2

Please give me a detailed ans i promise to rate every one

joyfrancis

hey is it sin (xlogcoscx) or sinx(logcosx)?

neeraj_agarwal_1990

plz use proper brackets...

waterdemon

You have given in the question

Y = Sinx Log Cosx

dy/dx = ?

I tried a lot and am not gettin the answer so

I suppose that the question must have been

y = Log _Cosx Sinx

y = Log Sinx / Log Cosx

By applying the division rule we get

For f(x) / g(x)

dy/dx = f ' (x) . g(x) - g ' (x) . f (x) / g (x)²

So We will get

dy / dx = [ Cotx.logx + tan x . log sinx ] / [ Log cosx]²

Hence proved

Cheers !!!!!!!!!!@@@@@@@

plZ rate me for my efforts if I am correct.

Aatish

well I suppose the answer provided by....RAKESH.....IS WRONG.....
and it will be only correct when the question is changed to wat waterdemon has done.........so I think the question shud be checked once again...and put up in a proper manner....

magiclko

kindly clear the question!!!!

udbarguj

cosxlogcosx-sinxtanx

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