If F(x)=∫0x2cos(t) dtF(x) = \displaystyle \int_0^{x^2} \cos(t)\, dtF(x)=∫0x2cos(t)dt, then F′(x)=F'(x) =F′(x)=A2xsin(x2)2x \sin(x^2)2xsin(x2)Bcos(x2)\cos(x^2)cos(x2)Csin(x2)−1\sin(x^2) - 1sin(x2)−1D2xcos(x2)2x \cos(x^2)2xcos(x2)check_circleExplanationBy FTC and the chain rule, F′(x)=cos(x2)⋅2x=2xcos(x2)F'(x) = \cos(x^2)\cdot 2x = 2x\cos(x^2)F′(x)=cos(x2)⋅2x=2xcos(x2).