Differentiate with respect to `x`, `y = sin^2(2x+1)`

Explanation

Let AB be a normal chord where `A-=(at^(2),2at)andB-=(at_(1)^(2),2at_(1))`.
We have `t_(1)=-t-(2)/(t)`.
Now, `AB^(2)=[a(t^(2)-t_(1)^(2))]^(2)+4a^(2)(t-t_(1))^(2)`
`=a^(2)(t-t_(1))^(2)[(t+t_(1))^(2)+4]`
`=a^(2)(t+t+(2)/(t))^(2)((4)/(t^(2))+4)`
`=(16a^(2)(1+t^(2))^(3))/(t^(4))`
For minimum length of chord AB,
`(d(AB^(2)))/(dt)=16a^(2)((t^(4)[3(1+t^(2))^(2)*2t]-(1+t^(2))^(3)*4t^(3))/(t^(8)))`
`=32a^(2)(1+t^(2))^(2)((3t^(2)-2-2t^(2))/(t^(5)))`
`(32a^(2)(1+t^(2))^(2))/(t^(5))(t^(2)-2)`
For `(d(AB^(2)))/(dt)=0,t=sqrt(2)` for which `AB^(2)` is minimum.
Thus, `AB_(min)=(4a)/(2)(1+2)^(3//2)=6sqrt(3)a` units