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Grade: 12
        
find dy/dx when
y= (sin x)^cos x  + (cos x)^sin x
5 years ago

Answers : (1)

Rinkoo Gupta
askIITians Faculty
80 Points
							

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

Thanks & Regards

Rinkoo Gupta

AskIITians Faculty

5 years ago
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