Answer to Question #203420 in Calculus for Rajay Myrie

Question

Show that 𝑑𝑦 𝑑𝑥 = 𝑠𝑒𝑐2𝑥 given that 𝑦 = tan x

Answer 1

Given that y = tanx

y = sinx/cosx

We know that d/dx(u/v) =

[-u(dv/dx)+v(du/dx)]/v²

y' = [(cosx)(cosx)-(sinx)(-sinx)]/cos²x

y '= [cos²x + sin²x]/cos²x

y '= 1/cos²x = sec²x

Hence proved