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Answer 1 · 2016-09-04T06:33:46

#dy/dx=3/((1+3x)ln(1+3x))#.

Let #(1+3x)=t, and, ln(1+3x)=lnt=u#, so that,

#y=ln(ln(1+3x)=ln(lnt)=lnu#

Thus, #y# is a function of #u#, #u# of #t#, and #t# of #x#.

Accordingly, by The Chain Rule,

#dy/dx#=#(dy)/(du)##(du)/(dt)##(dt)/(dx)#.................................#(star)#

#y=lnu rArr dy/(du)=1/u..........................(1)#

#u=lnt rArr (du)/dt=1/t..............................(2)#

#t=1+3x rArr (dt)/dx=3..............................(3)#

Therefore, by #(1),(2),(3), and, (star)#, we get,

#dy/dx=1/u*1/t*3=3/(tu)#

#:. dy/dx=3/((1+3x)ln(1+3x))#.

Enjoy Maths.!