Last Activity: 10 Years ago
let y1=esinx
apply chain rule
dy/dx=esinx.cosx
let y2=tanxx
taking log on both sides
logy2=xlogtanx
diff.w.r.to x , we get
(1/y2)dy2/dx=x.(1/tanx)sec2x +logtanx.1
dy2/dx=tanxx[x/(sinxcosx)+logtanx]
hence dy/dx=esinx.cosx+tanxx[x/(sinxcosx)+logtanx]
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Rinkoo Gupta
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