Çözüldü Exponential Derivative

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GRE Mathematics Test Practice Book
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(Page 21, Question 20, Answers: Page 52)

g(0) = e^(2·0 + 1) = e^1 = e....(I)
lim (x → 0) { g[ g(x) ] - g(e) } / x = { g[ g(0) ] - g(e) } / 0 = [ g(e) - g(e) ] / 0 = 0 / 0 indetermination, so L'Hospital's Rule can be applied;
lim (x → 0) { g[ g(x) ] - g(e) } / x = lim (x → 0) g '(x)·g '[ g(x) ] / 1 = g '(0)·g '[ g(0) ]....(II)
g '(x) = 2e^(2x + 1)....(III) ⇒ g '(0) = 2e^(2·0 + 1) = 2e....(IV)
Using the value at (I) in (III), g '(e) = 2e^(2e + 1)....(V)
Substituting (IV) and (V) for the equation (II), 2e·2e^(2e + 1) = 4e^(2e + 1 + 1) = 4e^(2e + 2).