Determine the first second and third derivatives of sin2x^2

Answer 1

y = sin(2x^2)
y' = 4x cos(2x^2)
y" = 4cos(2x^2) - 16x^2 sin(2x^2)
y"' = -48x sin(2x^2) - 64x^3 cos(2x^2)

Answer 2

To find the derivatives of sin(2x^2), we can start by using the chain rule.

First derivative:
y = sin(2x^2)
y' = cos(2x^2) * d/dx (2x^2)
= cos(2x^2) * 4x
= 4x cos(2x^2)

Second derivative:
y' = 4x cos(2x^2)
y" = (4 cos(2x^2)) * d/dx (4x) + (4x) * d/dx (cos(2x^2))
= 4 cos(2x^2) * 4 + 4x * (-2sin(2x^2)) * d/dx (2x^2)
= 16 cos(2x^2) - 8x sin(2x^2)

Third derivative:
y" = 16 cos(2x^2) - 8x sin(2x^2)
y"' = (16 cos(2x^2) - 8x sin(2x^2)) * d/dx (2x^2) + (-8sin(2x^2)) * d/dx (2x^2) + (-8x) * d/dx (sin(2x^2))
= (16 cos(2x^2) - 8x sin(2x^2)) * 4x + (-8sin(2x^2)) * 4x + (-8x) * (2cos(2x^2)) * d/dx (2x^2)
= 64x^2 cos(2x^2) - 32x^2 sin(2x^2) - 16x cos(2x^2)
= -16x cos(2x^2) - 32x^2 sin(2x^2) + 64x^2 cos(2x^2)
= -16x cos(2x^2) - 32x^2 sin(2x^2) + 64x^2 cos(2x^2)
= -48x sin(2x^2) - 64x^3 cos(2x^2)

Therefore, the first derivative is y' = 4x cos(2x^2),
the second derivative is y" = 16 cos(2x^2) - 8x sin(2x^2),
and the third derivative is y"' = -48x sin(2x^2) - 64x^3 cos(2x^2).