differentiate x + 1/x from first principle Share with your friends Share 15 Varun.Rawat answered this Let y = x + 1x⇒y + ∆y = x+∆x + 1x+∆x⇒∆y = x+∆x + 1x+∆x - y⇒∆y = x+∆x + 1x+∆x - x - 1x⇒∆y = ∆x + 1x+∆x - 1x⇒∆y = ∆x + x-x-∆xxx+∆x⇒∆y∆x = ∆x∆x - ∆x∆x×1xx+∆x⇒∆y∆x = 1 - 1xx+∆x⇒lim∆x→0∆y∆x = lim∆x→01 - 1xx+∆x⇒dydx = 1 - 1x2 -13 View Full Answer
