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

Question

Evelyn Agard · Accepted Answer

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)#

Luke Alvarez · Answer

undefined To find the first and second derivatives of $ \ln(\ln(x^2)) $, we'll apply the chain rule and the derivative of the natural logarithm function.

First Derivative:

$ \frac{d}{dx}[\ln(\ln(x^2))] $

$ = \frac{1}{\ln(x^2)} \cdot \frac{1}{x^2} \cdot 2x $

$ = \frac{2}{x \ln(x^2)} $

Second Derivative:

To find the second derivative, we differentiate the expression we obtained for the first derivative.

$ \frac{d}{dx} \left[ \frac{2}{x \ln(x^2)} \right] $

Using the quotient rule:

$ = \frac{2 \cdot (-x \ln(x^2))' - 2x \ln(x^2)'}{(x \ln(x^2))^2} $

$ = \frac{2 \cdot (-1) \cdot \ln(x^2) - 2x \cdot \frac{1}{x^2} \cdot 2x}{(x \ln(x^2))^2} $

$ = \frac{-2\ln(x^2) - 4}{(x \ln(x^2))^2} $

$ = \frac{-2\ln(x^2) - 4}{(x \cdot 2\ln(x))^2} $

$ = \frac{-2\ln(x^2) - 4}{(2x\ln(x))^2} $

$ = \frac{-\ln(x^2) - 2}{(x\ln(x))^2} $

So, the second derivative of $ \ln(\ln(x^2)) $ is $ \frac{-\ln(x^2) - 2}{(x\ln(x))^2} $.