calculus higher derivative help?

anonymous

given that y= ln(sin^3 2x) show that 3 d^2y/ dx^2 +(dy/dx)^2 +36=0

please help me step by step

thank you

kb

Since y = ln(sin^3(2x)) = 3 ln(sin(2x)):

dy/dx = 3 * (1/sin(2x)) * 2 cos(2x) = 6 cot(2x), and

d^2y/dx^2 = -12 csc^2(2x).

So, 3 d^2y/dx^2 + (dy/dx)^2 + 36

= 3 * -12 csc^2(2x) + (6 cot(2x))^2 + 36

= -36 csc^2(2x) + 36 cot^2(2x)) + 36

= -36 csc^2(2x) + 36 (cot^2(2x) + 1)

= -36 csc^2(2x) + 36 csc^2(2x)

= 0.

I hope this helps!