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

Apr 4, 2016

Since $\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$, it implies that, together with use of the power rule, we get

$\frac{d}{\mathrm{dx}} \ln \left(\ln \left(\ln f \left(x\right)\right)\right) = \frac{1}{\ln \left(\ln f \left(x\right)\right)} \cdot \frac{1}{\ln f \left(x\right)} \cdot \frac{1}{f} \left(x\right) \cdot \frac{\mathrm{df}}{\mathrm{dx}}$

Apr 5, 2016

$g ' \left(x\right) = \frac{f ' \left(x\right)}{f \left(x\right) \cdot \ln \left(f \left(x\right)\right) \cdot \ln \left(\ln \left(f \left(x\right)\right)\right)}$

#### Explanation:

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

$\frac{d}{\mathrm{dx}} \ln \left(\textcolor{red}{u}\right) = \frac{1}{\textcolor{red}{u}} \cdot \frac{\textcolor{b l u e}{\mathrm{dc}} o l \mathmr{and} \left(red\right) u}{\textcolor{b l u e}{\mathrm{dx}}}$

We must start with the outermost $\ln \left(u\right)$ 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,

$g ' \left(x\right) = \frac{1}{\ln} \left(\ln \left(f \left(x\right)\right)\right) \cdot \frac{1}{\ln} \left(f \left(x\right)\right) \cdot \frac{d}{\mathrm{dx}} \ln \left(f \left(x\right)\right)$

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)

$= \frac{1}{f} \left(x\right) \cdot f ' \left(x\right)$

All together, we see that

$g ' \left(x\right) = \frac{1}{\ln} \left(\ln \left(f \left(x\right)\right)\right) \cdot \frac{1}{\ln} \left(f \left(x\right)\right) \cdot \frac{1}{f} \left(x\right) \cdot f ' \left(x\right)$

$= \frac{f ' \left(x\right)}{f \left(x\right) \cdot \ln \left(f \left(x\right)\right) \cdot \ln \left(\ln \left(f \left(x\right)\right)\right)}$