f(x)=x²×(tgx+1)
f'(x)=(x²)'×(tgx+1)+ x²×(tgx+1)'=
2x×(tgx+1)+x²/cos²x
f''(x)=(2x×(tgx+1)+x²/cos²x)'=
2×(tgx+1)+2x/cos²x+(2x×cos²x-x²×2cosx×(-sinx))/(cosx)^4=
2×(tgx+1)+2x/cos²x+(2x×cos²x+x²×sin2x)/(cosx)^4
Ответ: f"(x)=2×(tgx+1)+2x/cos²x+(2x×cos²x+x²×sin2x)/(cosx)^4