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

2 Answers
Nov 6, 2015

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

Explanation:

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

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

By the chain rule

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

Nov 6, 2015

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

Explanation:

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