Differentiate (frac{x}{cosx}) with respect to x

A.
1 + x sec x tan x 
B.
1 + sec^{2} x 
C.
cos x + x tan x 
D.
x sec x tan x + secx
Correct Answer: Option D
Explanation
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