y^'=\(\dfrac{1}{cos^2x}-\dfrac{1}{cos^2x}tan^2x+\dfrac{1}{cos^2x}tan^4x\)
=\(\dfrac{1}{cos^2x}-\dfrac{sin^2x}{\cos^4x}+\dfrac{\sin^4x}{\cos^8x}\)
=\(\dfrac{\cos^4x-\sin^2.\cos^2x+\sin^4x}{\cos^8x}\)

=\(\dfrac{\left(\cos^2x+\sin^2x\right)^2-3\sin^2x\cos^2x}{\cos^8x}\)

=\(\dfrac{-3\sin^2x}{\cos^6x}\)