Does anyone know how to find the derivative of y= ln(lnx^2)?I tried to do everything, but I don't know how to start it. Please help me.

y= ln(lnx^2)

To find the derivative of the y.

We use thr results :  d/dx(x^n) = nx^(n-1) and

d/dx { f (g(x))} ={ f(g(x)}' = { f '(g(x))}*g'(x), the chain rule formula.

Therefore d/dx{ln(f(x)} ={ lnf(x)}'  = {1/f(x)}f'(x).

Coming to our question,

y = ln(lnx^2). To find the dy/dx:

dy/dx = (ln(lnx^2))'. Using the above chain rule formula we get:

dy/dx =  {1/(lnx^2)} (lnx^2)'

= {1/(lnx^2)} (1/x^2)(x^2)'

=(1/(lnx^2)(1/x^2)(2x)

=2/{xlnx^2)

=1/(xlnx)