How do I find the derivative of $g(x)=ln(ln(ln(f(x))))$ ?

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Answer 1 · 2016-04-04T21:32:40

Since $d/dxlnx=1/x$ , it implies that, together with use of the power rule, we get

$d/dxln(ln(lnf(x)))=1/(ln(lnf(x))) * 1/(lnf(x)) * 1/f(x) * (df)/dx$

Answer 2 · 2016-04-05T20:10:33

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

The chain rule, as it applies to the function $ln(x)$ , is:

$d/dxln(color(red)u)=1/color(red)u*(color(blue)dcolor(red)u)/color(blue)dx$

We must start with the outermost $ln(u)$ function:

$g'(x)=d/dxln(color(red)ln(ln(f(x))))=1/color(red)ln(ln(f(x)))*color(blue)(d/dx)color(red)(ln(ln(f(x)))$

Reapplying the rule to the new derivative, we see that

$d/dxln(color(red)ln(f(x)))=1/color(red)ln(f(x))*color(blue)(d/dx)color(red)(ln(f(x))$

Thus,

$g'(x)=1/ln(ln(f(x)))*1/ln(f(x))*d/dxln(f(x))$

For the final time, find a last natural logarithm derivative:

$d/dxln(color(red)f(x))=1/color(red)f(x)*color(blue)(d/dx)color(red)(f(x)$

$=1/f(x)*f'(x)$

All together, we see that

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

$=(f'(x))/(f(x)*ln(f(x))*ln(ln(f(x))))$

How do I find the derivative of g(x)=ln(ln(ln(f(x))))g(x)=ln(ln(ln(f(x))))?