How to find dy/dx if y=ln(sin(3x^2-1))*ln(cos(3x^2-1))?

To determine dy/dx, we'll have to use two rules of
differentiation: chain rule and product rule.

The product
rule states:

(f*g)' = f'*g +
f*g'

The chain rule
states;

[f(g(x))]' =
f'(g(x))*g'(x)

We'll write the product rule
first:

dy/dx = [ln(sin(3x^2-1))]'*ln(cos(3x^2-1)) +
ln(sin(3x^2-1))*[ln(cos(3x^2-1))]'

Now, we'll apply the
chain rule:

dy/dx =
6x*cos(3x^2-1)*ln(cos(3x^2-1))/(sin(3x^2-1) - 
6x*sin(3x^2-1)*ln(sin(3x^2-1))/cos(3x^2-1)

dy/dx =
6x*{[cos(3x^2-1)]^2*ln(cos(3x^2-1)) -
[sin(3x^2-1)]^2*ln(sin(3x^2-1))}/sin(3x^2-1)*cos(3x^2-1)

dy/dx
= 12x*{[cos(3x^2-1)]^2*ln(cos(3x^2-1)) -
[sin(3x^2-1)]^2*ln(sin(3x^2-1))}/2*sin(3x^2-1)*cos(3x^2-1)

We've
created at denominator, the formula of sine of double
angle:

dy/dx =
12x*{[cos(3x^2-1)]^2*ln(cos(3x^2-1)) - [sin(3x^2-1)]^2*ln(sin(3x^2-1))}/ sin (6x^2 -
2)