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

1 Answer
Nov 5, 2016

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.

Explanation:

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)