Find d^2y/dx^2 : x^2+y^2=1

Question

Find d²y/dx² : x²+y²=1

Answer 1

To get dy/dx, do implicit derivation

x²+y²=1

2x + 2y dy/dx = 0

dy/dx = -x/y

Now take another derivative

d²y/dx² = -(y - xdy/dx)/y² --> dy/dx = -x/y

= (-y + x(-x/y))/y²

= (-y -x²/y)/y²

=(-y²-x²)/y³ But we know x²+y² = 1

= -1/y³

Note: If y = √(1-x²)

d²y/dx² = -1/(√(1-x²))³ = -1/((1-x²)√(1-x²))

Answer 2

x² + y² = 1

2x + 2y(dy/dx) = 0

So, dy/dx = -x/y

d²y / dx² = [-y + x(dy/dx)] / y² = [-y + x(-x/y)] / y² = -1/y - x² / y³