Differentiation 2

Subparagraph before solution:

( This is not absolutely formal, but it works) Let's suppose $f: \mathbb{R} \longrightarrow \mathbb{R}$ to be a derivative function with continuous derivative fulfilling the hypothesis f(1) = 2 and f '(x) = f(x)

Let's call $f(x) = y \Rightarrow y ' = y \Rightarrow \dfrac{dy}{dx} = y \Rightarrow \dfrac{dy}{y} = dx$ $\Rightarrow \int \dfrac{dy}{y} = \int dx \Rightarrow \ln |y| = x + C \text { (where C is a constant)}$ . Now, applying exponential $y = f(x) = ke^x \text{ (k is a constant) }; \quad f(1) = y(1) = 2 = k \cdot e \Rightarrow k = \frac{e}{2}$ $\rightarrow f(x) = y = \frac{2}{e} \cdot e^x = 2 \cdot e^{x -1} \Rightarrow \ln(f(2015)) = \ln (2 \cdot e^{2014}) =$ $= \ln 2 + 2014$

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Solution.- I have just seen $f: \mathbb{R} \longrightarrow \mathbb{Z}$ . If f is a continuous function and differentiable function on $\mathbb{Z}$ then f is a constant function and then $0 = f '(x) = f(x) = 2 \space \forall x \in \mathbb{R}$ is impossible = contradiction, and if f is not a continuous function then there doesn't exist its derivative .... so this function does not exist