Find derivative of tan3x by first principle. Share with your friends Share 2 Varun.Rawat answered this Let y = tan 3x⇒y + ∆y = tan 3x + ∆x⇒y + ∆y = tan3x + 3∆x⇒∆y = tan3x + 3∆x - tan 3x⇒∆y = sin3x + 3∆xcos3x + 3∆x - sin 3xcos 3x⇒∆y = sin3x + 3∆x . cos 3x - cos3x + 3∆x . sin 3xcos3x + 3∆x . cos 3x⇒∆y = sin3x + 3∆x - 3xcos3x + 3∆x . cos 3x⇒∆y = sin 3∆xcos3x + 3∆x . cos 3x⇒∆y∆x = sin 3∆x3∆x×3×1cos3x + 3∆x . cos 3x⇒lim∆x→0∆y∆x = 3lim∆x→0sin 3∆x3∆x×lim∆x→01cos3x + 3∆x . cos 3x⇒dydx = 3 × 1 × 1cos 3x . cos 3x⇒dydx = 3 sec23x 1 View Full Answer Hafsa answered this Refer example 17 in NCERT in pg no.311 it is a similar sum. 0