To determine $g^{\prime}(0)$, we start by applying the chain rule and product rule to find the derivative of the given function $g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$. The product rule states that if we have two functions $u(x)$ and $v(x)$, then the derivative of their product is given by: $(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)$ Let's denote $u(x) = h(\mathrm{e}^x)$ and $v(x) = \mathrm{e}^{h(x)}$. First, we need to find $u'(x)$ and $v'(x)$. Using the chain rule, we find: $u'(x) = \frac{d}{dx}h(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \frac{d}{dx}(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \mathrm{e}^x$ Now, the derivative of $v(x)$ is: $v'(x) = \frac{d}{dx}(\mathrm{e}^{h(x)}) = \mathrm{e}^{h(x)} \cdot h'(x)$ Using the product rule, we get the derivative of $g(x)$: $g'(x) = u'(x) v(x) + u(x) v'(x)$ Substituting $u(x)$, $v(x)$, $u'(x)$, and $v'(x)$ into the above expression, we get: $g'(x) = \left( h'(\mathrm{e}^x) \mathrm{e}^x \right) \mathrm{e}^{h(x)} + \left( h(\mathrm{e}^x) \right) \left( \mathrm{e}^{h(x)} h'(x) \right)$ Next, we need to evaluate this at $x = 0$: First, we know that: $h(0) = 0$ $h(1) = 1$ $h'(0) = 2$ $h'(1) = 2$ Substituting $x = 0$ into the expressions, we get: $u(0) = h(\mathrm{e}^0) = h(1) = 1$ $v(0) = \mathrm{e}^{h(0)} = \mathrm{e}^0 = 1$ $u'(0) = h'(\mathrm{e}^0) \mathrm{e}^0 = h'(1) \cdot 1 = 2$ $v'(0) = \mathrm{e}^{h(0)} h'(0) = \mathrm{e}^0 \cdot 2 = 2$ Therefore, evaluating $g'(0)$: $g'(0) = \left( u'(0) v(0) \right) + \left( u(0) v'(0) \right)$ $g'(0) = \left( 2 \cdot 1 \right) + \left( 1 \cdot 2 \right)$ $g'(0) = 2 + 2 = 4$ Thus, the value of $g^{\prime}(0)$ is 4, which corresponds to Option A. The correct answer is Option A: 4 .