natural logarithms

Question

Let f(x) = ln(ln(x)). Find f' (e⁴)

Answer 1

You need to use the chain rule, (f(g(x)))' = f'(g(x))*g'(x) and the fact that ln(x)' = 1/x.

Then, as f(x) = ln(x), and g(x) = ln(x), we have f'(g(x)) = 1/ln(x) and g'(x) = 1/x, so

f'(g(x))*g'(x) = 1/ln(x) * 1/x = 1/(x ln(x) )

Then, at e^4, ln(e^4) = 4, so

1/(x ln(x) ) = 1/(e^4*4) = 1/(4e^4) or e^-4/4

 

Answer 2

Another way to do it is by implicit differentiation. 

 

e^y=e^ln(ln(x))

 

differentiate implicitly

 

e^y y'= 1/x

 

you know y = ln(lnx)

 

e and ln cancel each other out

 

so you get e^ln(lnx) y'=1/x

 

you will in turn get

 

ln (x) y' = 1/x

 

Divide my ln (x) and you will get

 

y'=1/(x ln(x))

 

Then do exactly as above and plug in the value.  You will get the same answer by using implicit differentiation.