How do you find the first and second derivative of #ln(lnx^2)#?

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Answer 1 · 2016-08-18T13:32:06

#(1) : dy/dx=1/(xlnx)#.

#(2) : (d^2y)/dx^2=-(1+lnx)/(x^2(lnx)^2)#.

Let #y=ln(lnx^2)#

Using the well-known Rules of Log. Fun., we have,

#y=ln(2lnx)=ln2+ln(lnx)#

By the Chain Rule, then,

#dy/dx=d/dx(ln2)+d/dx{ln(lnx)}#

#=0+1/lnx*d/dx(lnx)#

#:. dy/dx=1/lnx*1/x=1/(xlnx)#.

Before proceeding further for the #2^(nd)# Derivative #(d^2y)/dx^2#, let us recall that #d/dt(1/t)=-1/t^2#. Hence,

#(d^2y)/dx^2=d/dx(dy/dx)=d/dx{1/(xlnx)}#

#=-1/((xlnx)^2)*d/dx{xlnx}#

#=-1/(x^2(lnx)^2){x*d/dx(lnx)+(lnx)*d/dx(x)}#

#=-1/(x^2(lnx)^2){x*1/x+(lnx)(1)}#

#:. (d^2y)/dx^2=-(1+lnx)/(x^2(lnx)^2)#.

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