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This must be a simple answer, yet I am unsure because of the fact that it has a precise interval.

Would the derivative of $g(x)=\int^{x}_{0}e^{\frac{t^2}{2}}dt$ be simply: $$g'(x)= e^{\frac{x^2}{2}}$$ ?

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  • $\begingroup$ Yes. You can use the fundamental theorem of calc here. $\endgroup$
    – Chinny84
    Apr 4, 2017 at 15:36
  • $\begingroup$ But of course that is the answer. $\endgroup$ Apr 4, 2017 at 15:36

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In general if : $$g(x)=\int^{x}_{a}f(t)dt$$ Then $$g'x)=f(x)$$ Indeed let $F$ an antiderivative of $f$ then with fundamental theorem of calculus : $$g(x)=F(x)-F(a)\Rightarrow g'(x)=F'(x)=f(x)$$

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Note that the function $f: t \mapsto e^{t^{2}/2}$ is continuous everywhere on $\mathbb{R}$; so by the fundamental calculus theorem, if $0 \in ]a,b[$ and if the function $F: x \mapsto \int_{0}^{x}f$ is defined on $]a,b[$, then $F$ is a primitive of $f$ on $]a,b[$. So $DF(x) = f(x) = e^{x^{2}/2}$.

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