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let y1=(sinx)cosx
taking log on both sides
logy1=cosxlogsinx
diff. w.r. to x
1/y1.dy1/dx=cosx.(1/sinx).cosx+logsinx.(-sinx)
dy1/dx=(sinx)cosx[cos2x- sin2x.logsinx]/sinx
lety2=(cosx)sinx
logy2=sinxlogcosx
diff.w.r.to x
(1/y2)dy2/dx=sinx.(1/cosx)(-sinx)+logcosx .cosx
dy2/dx=(cosx)sinx[-sin2x+cos2x. logcosx]/cosx
dy/dx= sinx)cosx[cos2x- sin2x.logsinx]/sinx
+=(cosx)sinx[- sin2x+cos2x. logcosx]/cosx
Thanks & Regards
Rinkoo Gupta
AskIITians Faculty
ogy1=cosxlogsinx
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