Differentiate `(x^2(1-x^2)^3)/(cos2x)` with respect to `x` .

Explanation

Let `y=(x^2(1-x^2)^3)/(cos2x)`
`(dy)/(dx)=d/(dx)(x^2(1-x^2)^3)/(cos2x)`
`=(d/(dx)[x^2(1-x^2)^3]*cos 2x -x^2(1-x^2)^3*d/(dx) (cos2x))/cos^2 (2x)`
`=([2x(1-x^2)^3+x^2*3(1-x^2)^2-2x]cos2x-(x^2)(1-x^2)^3-sin 2x*2)/ cos^2 (2x)`
`=([2x (1-x^2)^3-6x63(1-x^2)^2]cos 2x + 2x^2 (1-x^2)63 sin 2x)/cos^2(2x)`
`=(x(1-x^2)^2)/cos^2 (2x)[{2(1-x^2)-6x^2}cos2x+2x(1-x^2)sin 2x ]`
`=(x(1-x^2)^2)/cos^2 (2x)[(2-8x^2)cos 2x +2x (1-x^2) sin 2x ]`