How do you find the first and second derivative of #ln(ln x^2)#?

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Answer 1 · 2016-11-05T12:50:44

For the first, using chain rule, which states that #(dy)/(dx)=(dy)/(dv)(dv)/(du)(du)/(dx)#
For the second, using quotient and product rules.

Renaming #v=lnu# and #u=x^2#, we get that

#(dy)/(dx)=1/v1/u2x=(2x)/(u*lnu)=(2cancel(x))/(x^cancel(2)*lnx^2)#

For the second derivative, we must recall quotient rule...

#(f(x)/g(x))'=(f'g-fg')/g^2#

...and product rule (to derive the quotient):

#(a(x)b(x))'=a'b+ab'#

Thus:

#(dy^2)/(dx^2)=(cancel(0*(xlnx^2))-2(lnx^2+1))/(xlnx^2)^2#

#(dy^2)/(dx^2)=-(2lnx^2+2)/(x^2*(lnx^2)^2)#