For g(x)=∫2exlnt dtg(x) = \int_2^{e^x} \ln t\,dtg(x)=∫2exlntdt, g′(x)=g'(x) =g′(x)=Aln(ex)\ln(e^x)ln(ex)Bexln(ex)=xexe^x \ln(e^x) = x e^xexln(ex)=xexcheck_circleCxexx e^xxexDx⋅exx \cdot e^xx⋅exExplanationChain rule: ln(ex)⋅ex=xex\ln(e^x) \cdot e^x = x e^xln(ex)⋅ex=xex.