dy/dx=y [ √ 1 − y ^2 ]. find d^2y/dx^2

Question

im having slight trouble with this one

Answer 1

dy/dx = y√(1-y²)

d²y / dx² = (dy/dx)√(1-y²) + y(1/2)(1-y²)^-1/2(-2y)(dy/dx)

= (dy/dx) [ √(1-y²) - y² / √(1-y²) ]

= (dy/dx) [ (1 - 2y²) / √(1-y²)

= y√(1-y²) [ (1 - 2y²) / √(1-y²)

= y(1-2y²) = y - 2y³