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v/Find dy.when y =(sin xyosx+(COSX)smxcoSX |
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Answer» 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 taking log on both sides 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 |
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