MathematicsHard63×since 2002Q3661Suppose for a differentiable function h,h(0)=0,h(1)=1h, h(0)=0, h(1)=1h,h(0)=0,h(1)=1 and h′(0)=h′(1)=2h^{\prime}(0)=h^{\prime}(1)=2h′(0)=h′(1)=2. If g(x)=h(ex)eh(x)g(x)=h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}g(x)=h(ex)eh(x), then g′(0)g^{\prime}(0)g′(0) is equal to:A4B5C3D8Check answerSkip