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Find dy/dx if y+siny=cosx...

Find `dy/dx if y+siny=cosx`

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The correct Answer is:
A, C

`S_(n)(x)=(x^(n-1)+(1)/(x^(n-1)))+2(x^(n-2)+(1)/(x^(n-2)))+"....."+(n-1)(x+(1)/(x))+n`
Let `S'=x^(n-1)+2x^(n-2)+3x^(n-3)+"....."+(n-1)x`
`((S')/(x)=x^(n-2)+2x^(n-3)+"....."+(n-2)x+(n-1))/(S'(1-(1)/(x))=x^(n-1)+x^(n-2)+x^(n-3)+"....."+x-(n-1))`
`S'((x-1))/(x)=(x*x^(n-1)-1)/((x-1))-(n-1)`
`implies S'=(x^(2))/((x-1)^(2))(x^(n-1)-1)-((x-1)x)/((x-1))`
` S''=(1)/(x^(n-1))+(2)/(x^(n-2))+"....."+((n-1))/(x)`
`implies S''=(1)/(x^(n))[x+2x^(2)+"......."+(n-1)x^(n-1)]`
` =(1)/(x^(n))([(n-1)x^(n)-nx^(n-1)+1])/((x-1)^(2))[" similarly as above "]`
`:.S_(n)(x)=S'+S''+n`
`implies S_(n)(x)=(1)/(x^((n-1)))((x^(n)-1)/(x-1))^(2)" " "......(i)"`
For`n=1,S_(1)(x)=1`
`S_(100)(x)=(1)/(x^(99))((x^(100)-1)/(x-1))^(2)`.
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