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

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Answer 1 · 2016-01-01T20:29:41

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

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

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

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

Substituting #u#, then #v# and finally #w#:

#(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))#