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sin^(n)(ax^(2)+bx+c)...

`sin^(n)(ax^(2)+bx+c)`

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The correct Answer is:
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Let `y = sin^(n)(ax^(2)+bx+c)`
`:. (dy)/(ddx) = (d)/(dx)[sin(ax^(2)+bx+c)]^(n)`
`= n.[sin(ax^(2)+bx+c)]^(n-1).(d)/(dx)sin(ax^(2)+bx+c)`
`= n.sin^(n-1)(ax^(2)+bx+x).cos(ax^(2)+bx+c).(2ax+b)`
`= n. (2ax+b).sin^(n-1)(ax^(2)+bx+c).cos(ax^(2)+bx+c)`
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