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let y1=(sinx)cosx
taking log on both sides
logy1=cosx.logsinx
diff. w. r. to x using product rule,we get
1/y1.dy1/dx=cosx.(1/sinx).cosx+logsinx.(-sinx)
dy1/dx=(sinx)cosx[cos2x- sin2x.logsinx]/sinx
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Rinkoo Gupta
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