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Let y1=(sinx)sinx
taking log on both sides, get
logy1=sinx . log (sinx)
diff.w.r.to x , we get
(1/y1).dy1/dx=sinx.(1/sinx).cosx+log(sinx).cosx
=cosx+cosx.log(sinx)
dy1/dx=(sinx)sinx[cosx+cosx.log(sinx)]
=(sinx)sinx.cosx[1+log(sinx)]
let y2=sin-1√x
dy2/dx=(1/√(1-x)).(1/2√x)
=(1/2).√x(1-x)
Hence dy/dx=dy1/dx+dy2/dx
dy/dx =(sinx)sinx.cosx[1+log(sinx)]+1/[2√{x(1-x)}] Ans.
Thanks & Regards
Rinkoo Gupta
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