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What is the $2^{nd}$ derivative of the normal distribution at one standard deviation?

The normal distribution is given by $N(x,\mu ,\sigma)=\frac{1}{\sigma\sqrt{2\pi }}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$. To make this problem easier, lets say I have a standard normal distribution($\mu =0,\sigma =1$). So $N\left(x,0,1\right)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$.

$$\frac{d}{dx}N(x,0,1)=\frac{d}{dx}\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\right)$$

$$\frac{d}{dx}N\left(x,0,1\right)=\frac{1}{\:\sqrt{2\pi}}\frac{d}{dx}\left(e^{-\frac{x^2}{2}}\right)$$

$$\frac{d}{dx}N\left(x,0,1\right)=\frac{1}{\:\sqrt{2\pi}}e^{-\frac{\left(x\:\right)^2}{2}\cdot \frac{d}{dx}\left(-\frac{x^2}{2}\right)}$$

$$\frac{d}{dx}N\left(x,0,1\right)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}\cdot -x}$$

So, $$\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}\cdot\:-x}$$ is the first derivative. To get the second, I took the derivative of the first.

$$\frac{d}{dx}\left(\frac{d}{dx}N\left(x,0,1\right)\right)=\frac{d}{dx}\left(\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}\cdot \:-x}\right)$$

$$\frac{d}{dx}\left(\frac{d}{dx}N\left(x,0,1\right)\right)=\frac{1}{\:\sqrt{2\pi }}\frac{d}{dx}\left(e^{-\frac{x^2}{2}\cdot\:-x}\right)$$

$$\frac{d}{dx}\left(\frac{d}{dx}N(x,0,1)\right)=\frac{1}{\:\sqrt{2\pi}}e^{-\frac{x^2}{2}\cdot\:-x}\cdot \frac{d}{dx\:}\left(-\frac{x^2}{2}\cdot \:-x\right)$$

$$\frac{d}{dx}\left(\frac{d}{dx}N\left(x,0,1\right)\right)=\frac{1}{\sqrt{2\pi}}\left(e^{-\frac{(x^2}{2}\cdot \:-x}\right)\cdot \left(3\frac{x^2}{2}\right)$$

So now I evaluate $$\frac{1}{\sqrt{2\pi}}\left(e^{-\frac{x^2}{2}\cdot -x}\right)\cdot \left(3\cdot \frac{x^2}{2}\right),$$ at the standard deviation which I set to $\sigma =1$

So, $$\frac{1}{\sqrt{2\cdot :\pi}}\left(e^{-\frac{1^2}{2}\cdot -1}\right)\cdot \left(3\cdot\frac{1^2}{2}\right)$$

$$\frac{1}{\sqrt{2\pi}}\left(e^{-\frac{(1)^2}{2}\cdot\,-1}\right)\cdot \left(3\cdot \frac{1^2}{2}\right)=\frac{3\sqrt{e}}{2\sqrt{2\cdot\pi}}$$

But my teacher says the answer is suppose to be $0$? What am I doing wrong? Side Note: I am new to Calculus. So an elaborate explanation will be appreciated.

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    $\begingroup$ You are not differentiating correctly. $\endgroup$
    – uniquesolution
    Oct 17, 2015 at 22:26
  • $\begingroup$ @uniquesolution Can you elaborate? I am new to calculus and just recently learned the rules. $\endgroup$
    – James Smith
    Oct 17, 2015 at 22:27
  • $\begingroup$ First, you are dealing with the density functions of the general and standard normal distributions. Second, @MichaelHardy has shown you how to use the chain rule to get the first derivative. Notice that it is 0 for $x = 0$, which corresponds to a maximum of the density function (called a 'mode' in probability) at $x=0.$ Third, if you look at a graph of the std normal density function you will see 'inflection points' at $\pm 1$. Do you know the connection between second derivatives and inflection points? $\endgroup$
    – BruceET
    Oct 17, 2015 at 22:33
  • $\begingroup$ Now that you have edited to recognize the Answer, a second error in your differentiation appears: You should have two terms in your second derivative. (Look up derivative of a product.) $\endgroup$
    – BruceET
    Oct 17, 2015 at 22:36

1 Answer 1

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Your way of applying the chain rule is wrong. Here's the right way: $$ \frac d {dx} e^{-x^2/2} = e^{-x^2/2} \cdot \frac d {dx} \left( \frac{-x^2} 2 \right) = e^{-x^2/2} \cdot(-x). $$

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  • $\begingroup$ I thought the chain rule was derivative of the outside at the inside multiplied by the derivative of the inside.? $\endgroup$
    – James Smith
    Oct 17, 2015 at 22:28
  • $\begingroup$ Wait, I see that exactly what you said. So am I suppose to multiply within the exponent or within the statement? $\endgroup$
    – James Smith
    Oct 17, 2015 at 22:30
  • $\begingroup$ This is the chain rule: $\displaystyle \frac d {dx} f(g(x)) = f'(g(x)) \cdot g'(x)$. The following is wrong: $\displaystyle \frac d {dx} f(g(x)) = f'\Big(g(x) \cdot g'(x)\Big)$. The latter is what you did, with $f(x) = e^x$ and $g(x) = -x^2/2$. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Oct 17, 2015 at 22:30
  • $\begingroup$ Oh, I see. Thanks. I get it now. $\endgroup$
    – James Smith
    Oct 17, 2015 at 22:31
  • $\begingroup$ "$\{$derivative of the outside at the inside$\}$ multiplied by the derivative of the inside" is correct. $\qquad$ "derivative of the outside at $\{$the inside multiplied by the derivative of the inside$\}$" is not. ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Oct 17, 2015 at 22:33

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