Question

solve this logarithmic differentiation
If don't know don't answer plzz
​

Answer 1

Take the logarithm of the given function: lny=ln(xcosx),⇒lny=cosxlnx. Differentiating the last equation with respect to x, we obtain: (lny)′=(cosxlnx)′,⇒1y⋅y′=(cosx)′lnx+cosx(lnx)′,⇒y′y=(−sinx)⋅lnx+cosx⋅1x,⇒y′y=−sinxlnx+cosxx,⇒y′=y(cosxx−sinxlnx).

Answer 2

hope this attachment will help you

mark as brainlest answer please