How do you differentiate #ln(xy)=cos(y^4)#?

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Answer 1 · 2016-11-09T17:04:44

#(dy)/(dx)=-y/(x(4y^4sin(y^4)+1)#

#ln(xy)=cos(y^4)#

#lnx+lny=cos(y^4)#

differentiate wrt #x#

#d/(dx)(lnx+lny)=d/(dx)(cosy^4)#

#1/x+1/y(dy)/(dx)=4y^3(dy)/(dx)(-sin(y^4))#

#4y^3(dy)/(dx)sin(y^4)+1/y(dy)/(dx)=-1/x#

#(dy)/(dx)(4y^3sin(y^4)+1/y)=-1/x#

#(dy)/(dx)((4y^4sin(y^4)+1)/y)=-1/x#

#(dy)/(dx)=-y/(x(4y^4sin(y^4)+1)#