# Proving the continuity of the first derivative of the standard normal

Question: Let $\phi(x)=\frac{d}{dx}\left((1/\sqrt{2\pi})\exp(-x^2/2)\right)$, show that $\phi$ is continuous. (Edit: Corrected minor error here.)

Answer: I'm not very familiar with analysis (and have little experience with proofs), but as I recall, the conditions of continuity for $\phi$ is that for any $x,y\in\mathbb R$, $\forall\epsilon_{>0}\exists\delta_{>0}$ s.t. $|f(x)-f(y)|<\epsilon\Rightarrow |x-y|<\delta$. Hence:

Attempt: For some $x,y\in\mathbb R$ s.t. $x<y$ (without loss of generality), suppose $|\phi(x)-\phi(y)|<\epsilon$. Then

$$\left|\min_{s\in[x,y]}\{\phi'(s)\}\right|\leq\frac{|\phi(x)-\phi(y)|}{|x-y|}<\frac\epsilon{|x-y|}$$

Thus,

$$|x-y|<\frac{\epsilon}{\left|\min_{s\in[x,y]}\{\phi'(s)\}\right|}$$

Thus, for any $\epsilon$, choose $\delta$ to be the RHS.

Is this attempt rigorous, correct, or am I wrong and/or missing steps?

The function $$\phi(x)=\frac{d}{dx}(\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}})= -\frac{x}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$$ is continuous as a product of two continuous functions $x$ and $\frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$. The fact that you suggested\allowed to differentiate the later one assumes its continuity as given.