find the derivative by 1st principle xroot x Share with your friends Share 0 Varun.Rawat answered this Let y = xx = x3/2⇒y + ∆y = x + ∆x3/2⇒∆y = x + ∆x3/2 - y⇒∆y = x + ∆x3/2 - x3/2⇒∆y = x+∆x3 - x3⇒∆y∆x = x+∆x3 - x3∆x⇒∆y∆x = x+∆x3 - x3x + ∆x - x⇒∆y∆x =x+∆x - xx + ∆x + x + x2 + x∆xx+∆x - xx+∆x + x⇒∆y∆x = 2x + ∆x + x2 + x∆xx+∆x + x⇒lim∆x→0∆y∆x = lim∆x→02x + ∆x + x2 + x∆xx+∆x + x⇒dydx = 2x + 0 + x2 + 0x + x⇒dydx = 3x2x⇒dydx = 32x 0 View Full Answer Himank Garg answered this I hope it help:-) 0
