यदि (If) `y=sin^(n)(ax^(2)+bx+c)`, (find) `(dy)/(dx)` ज्ञात करें ।

Explanation

माना कि `y=sin^(n)(ax^(2)+bx+c)`
माना कि `u=ax^(2)+bx+c,v=sinu,` तो `y=v^(n)`
`therefore(du)/(dx)=2ax+b,(dv)/(du)=cosu` तथा `(dy)/(dv)=nv^(n-1)`
chain rule से , `(dy)/(dx)=(dy)/(dv)xx(dv)/(du)xx(du)/(dx)=nv^(n-1)*cosu*(2ax+b)`
`=nsin^(n-1)(ax^(2)+bx+c)cos(ax^(2)+bx+c)(2ax+b)`
`=n(2ax+b)sin^(n-1)(ax^(2)+bx+c)cos(ax^(2)+bx+c)`
Second method : `(dy)/(dx)=(d)/(dx)[sin^(n)(ax^(2)+bx)+c)]`
`=(d)/(d sin(ax^(2)+bx+c))sin^(n)(ax^(2)+bx+c)*(d)/(d(ax^(2)+bx+c))sin(ax^(2)+bx+c)*(d)/(dx)(ax^(2)+bx+c)`
`=n sin^(n-1)(ax^(2)+bx+c)cos(ax^(2)+bx+c)(2ax+b)`