Question 72

Suppose for a differentiable function $$h$$, $$h(0) = 0$$, $$h(1) = 1$$ and $$h'(0) = h'(1) = 2$$. If $$g(x) = h(e^x)e^{h(x)}$$, then $$g'(0)$$ is equal to:

g(x) = h(eˣ)·e^{h(x)}. g'(x) = h'(eˣ)eˣ·e^{h(x)} + h(eˣ)·e^{h(x)}·h'(x).

At x=0: g'(0) = h'(1)·1·e^{h(0)} + h(1)·e^{h(0)}·h'(0) = 2·e⁰ + 1·e⁰·2 = 2+2 = 4.

The correct answer is Option (2): 4.

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