Correct Answer  Option 1 : x(xsec
2x + 2tan x)
Concept:

\(\rm d\over dx\)xn = nxn1

\(\rm d\over dx\)sin x = cos x

\(\rm d\over dx\)cos x = sin x

\(\rm d\over dx\)ex = ex

\(\rm d\over dx\)ln x = \(\rm1\over x\)

\(\rm d\over dx\)tan x = sec2 x

\(\rm d\over dx\)[f(x)g(x)] = g(x)\(\rm d\over dx\)f(x) + f(x)\(\rm d\over dx\)g(x)
Chain Rule: If y is a function of u and u is a function of x
 \(\rm {dy\over dx} = {dy\over du}\times {du\over dx}\)
Calculation:
f(x) = x2 tan x
f'(x) = \(\rm d\over dx\)f(x) = \(\rm d\over dx\)(x2 tan x)
f'(x) = tan x \(\rm d\over dx\)(x2) + x2 \(\rm d\over dx\)tan x
f'(x) = tan x (2x) + x2 (sec^{2} x)
f'(x) = x(xsec2x + 2tan x)
∴ The required value is x(xsec2x + 2tan x).