What is the derivative of #f(x) = ln(lnx)#?

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Answer 1 · 2015-11-06T21:51:30

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

Let #y=lnu# then #(dy)/(du)=1/u#

Let #u=lnx# then #(du)/dx=1/x#

#(dy)/dx=(dy)/(du)(du)/dx=1/u(1/x)=1/lnx(1/x)=1/(xlnx)#

Answer 2 · 2015-11-06T21:52:24

The derivative is #1/(xln(x))#

You have a composite function #f(g(x))#, with the special case in which #f(x)=g(x)=ln(x)#

So, the chain rule states that

#(f(g(x))' = f'(g(x)) * g'(x)#,

and we need to use the fact that the derivative of #ln(x)# is #1/x#.

So, we have that #f'(g(x)) = 1/g(x)=1/ln(x)#, and #g'(x)=1/x#.

Putting everything together, we have

#d/dx ln(ln(x))= 1/ln(x) * 1/x = 1/(xln(x))#