find dy/dx...if y=e^log[log(logx)]*log3Xpls note log3x does not include in the exponent...help me pls...thanks in advance...

differentiation of e^x is e^x only

now differentiation of e^(log[log(logx) *log3x is

know uv rule

diff of u*v+diff ofv*u

so the diff is (1/3x)*3*e^m  +(log3x)*e^m*(dm/dx).........(1)                  let m=log(log(logx)))

dm/dx is equal to   (1/loglogx) * (1/logx) * (1/x)*1......

substitute dm/dx in (1) and get the answer

formulaes used=(1) differentiation of e^m is e^m *(dm/dx)

                        (2) differentiation of logn is (1/n) * (dn /dx0

hope u understood