Question

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

Answer 1

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

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

Explanation:

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))))`