What are the first and second derivatives of f(x)=ln(ln(ln(8x))) f(x)=ln(ln(ln(8x)))?

1 Answer
Jan 1, 2016

Using chain rule, which here will need to be made longer.

Explanation:

Let's rename u=ln(v)u=ln(v), v=ln(8x)v=ln(8x) and w=8xw=8x

Chain rule states that (dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dw)(dw)/(dx)dydx=dydududvdvdwdwdx

(dy)/(dx)=1/u1/v1/w(8)=8/(uvw)dydx=1u1v1w(8)=8uvw

Substituting uu, then vv and finally ww:

(dy)/(dx)=8/(ln(v)vw)=8/(ln(ln(w))ln(w)w)=cancel(8)/(ln(ln(8x))ln(8x)cancel(8)x)

(dy)/(dx)=1/(xln(8x)ln(ln(8x)))=(xln(8x)ln(ln(8x)))^-1

Now, we'll need chain rule and product rule for three terms. Besides, we'll need chain rule for two of the three terms.

For a product of three terms, the rule follows:
(abc)'=a'bc+ab'c+abc'

(dy^2)/(d^2x)=-u^-2[1ln(8x)ln(ln(8x))+cancel(x)(1/cancel(x))ln(ln(8x))+cancel(xln(8x))(1/(cancel(xln(8x))))]

Substituting u

(dy^2)/(d^2x)=-(ln(ln(8x))(ln(8x)+1)+1)/(xln(8x)ln(ln(8x)))^2

(dy^2)/(d^2x)=-(ln(ln(8x))(ln(8x)+1)+1)/(x^2ln^2(8x)ln^2(ln(8x))