What is d squared y over dx squared of the equation y = ln(x3) − sin(2x)?

Answer 1

y = ln(x³) - sin(2x)

dy/dx = (x³)'·(1/x³) - (2x)'cos(2x) <-- Use Chain Rule

dy/dx = 3x²/x³ - 2cos(2x)

dy/dx = 3/x - 2cos(2x)

d²y/dx² = (3/x)' - (2x)'[-2sin(2x)] <-- Use Chain Rule

d²y/dx² = -3/x² - 2[-2sin(2x)]

d²y/dx² = -3/x² + 4sin(2x)

Hope this helped!