1 + x sec x tan x
1 + sec2 x
cos x + x tan x
x sec x tan x + secx
Correct answer is D
let y = \(\frac{x}{cosx}\) = x sec x
y = u(x) v (x0
\(\frac{dy}{dx}\) = U\(\frac{dy}{dx}\) + V\(\frac{du}{dx}\)
dy x [secx tanx] + secx
x = x secx tanx + secx