If y=log(sinx)cosx, then dy/dx is equal ...

If `y=log_(sinx)cosx`, then dy/dx is equal to

A

`(tanxlog(sinx)-cotxlogcosx)/((logsinx)^2)`

B

`(-tanxlogsinx-cotxlogcosx)/((logsinx)^2)`

C

`(-tanxlog(sinx)-cotxlogcosx)/((logsinx)^2)`

D

`(tanxlog(sinx)-cotxlog(cosx))/([(logsinx)^2])

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The correct Answer is:

B

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