Find derivative of a natural log exponential function?

Question

Find dy/dx of:

y = (ln(x^2))^(2x+3)

Answer 1

y = (ln x²)^(2x+3) = e^((2x+3) ln ln x²)
Now just ruthlessly apply the chain rule:
dy/dx = e^((2x+3) ln ln x²) * (2 ln ln x² + (2x+3) * 1/ln x² * 1/x² * 2x)
And simplify:
(2 ln x)^(2x+3) * (2 ln (2 ln x) + (2x+3)/(x ln x))