Let `z=(cosx)hat5a n dy="sin"xdot` Then the value of `2(d^2z)/(dy^2)a tx=(2pi)/9` is____________.

Explanation

`z=(cos x)^(5),y=sin x`
`(dz)/(dx)=-5cos^(4)xcdot sin x, (dy)/(dx)= cos x`
`therefore" "(dz)/(dy)=-5cos^(3)x cdot sin x`
`"Now, "(d^(2)z)/(dy^(2))=(d)/(dx)((dz)/(dy))cdot(dx)/(dy)`
`=-5(d)/(dx)[cos^(3)xcdot sin x] (1)/(cos x)`
`=-5[cos^(4)x- 3 sin^(2)xcdot cos^(2)x ](1)/(cos x)`
`=-5(cos^(3)x-3 sin^(2)xcdotcos x)`
`=-5(cos^(3)x-3 cos x(1-cos^(2)x))`
`=-5(4 cos^(3)x-3 cos x)`
`=-5 cos 3x`
`therefore" "(d^(2)z)/(dy^(2)):|_(x=(2pi)/(9))=-5 cos 120^(@)=(5)/(2)`