x If G(x)=∫ax2f(t) dtG(x) = \int_a^{x^2} f(t)\,dtG(x)=∫ax2f(t)dt, then G′(x)G'(x)G′(x) equals:Af(x2)f(x^2)f(x2)Bf(x2)⋅2xf(x^2) \cdot 2xf(x2)⋅2xcheck_circleC2xf(x)2x f(x)2xf(x)Df(x)f(x)f(x)ExplanationBy FTC and chain rule, G′(x)=f(x2)⋅ddx(x2)=2xf(x2)G'(x) = f(x^2) \cdot \frac{d}{dx}(x^2) = 2x f(x^2)G′(x)=f(x2)⋅dxd(x2)=2xf(x2).