How do you find the derivative of #ln(ln x^2)#?

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Answer 1 · 2018-05-01T09:05:24

#1/(xlnx)#

We have:

#d/dx(ln(lnx^2))#

According to the chain rule, #d/dxf(g(x))=d/dx(f)*d/dxg(x)#

Here, #f(u)=lnu# where #u=lnx^2#

Since #d/(du)lnu=1/u#, we now have:

#1/u*d/dxlnx^2#

Here, #f(v)=lnv# where #v=x^2#, so we have:

#1/u*1/v*2x#

Since #u=lnx^2# and #v=x^2#, we have:

#(2x)/(ln(x^2)x^2)#

Since #lnx^n=nlnx#, we get:

#(2x)/(2x^2lnx)#

#1/(xlnx)#

Answer 2 · 2018-05-01T09:10:45

#d/dxln(lnx^2)=2/(xlnx^2)=1/(xlnx)#

We can use chain rule here. We can write #f(x)=ln(lnx^2)# as

#f(x)=ln(g(x))#, #g(x)=ln(h(x))# and #h(x)=x^2#

then #(df)/(dg)=1/(g(x))#, #(dg)/(dh)=1/(h(x))# and #(dg)/(dh)=2x#

and using chain rule as #(df)/(dx)=(df)/(dg)xx(dg)/(dh)xx(dh)/(dx)#

= #1/(g(x))xx1/(h(x))xx2x#

= #1/lnx^2*1/x^2*2x#

= #2/(xlnx^2)#

Hence #d/dxln(lnx^2)=2/(xlnx^2)=2/(x2lnx)=1/(xlnx)#