find dy/dx when

y= (sin x)^cos x + (cos x)^sin x

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