Derivative of $\frac{d}{dx}\{\exp \frac{-(x-\mu)^2)}{2 \sigma^2}\}$

Question

How to find derivative of this complex exponent?

$$\dfrac{d}{dx}\{\exp \dfrac{-(x-\mu)^2)}{2 \sigma^2}\}$$

$\frac{\mathrm d}{\mathrm dx}\mathrm e^{f(x)}=\mathrm e^{f(x)}\frac{\mathrm d}{\mathrm dx}f(x)$ — Nikolaj-K, Sep 24 '13 at 06:44

score 1 · Accepted Answer · answered Sep 24 '13 at 07:38

Note that $$\text{exp}{\left(\frac{-(x-\mu)^2}{2\sigma^2}\right)}=\text{exp}{\left(\frac{-x^2+2x\mu-\mu^2}{2\sigma^2}\right)}=\big(\text{exp}(-x^2)\times\text{exp}(+2x\mu)\times A\big)B$$ wherein $A=\text{exp}(-\mu^2),~~B=\text{exp}(-2\sigma^2)$ are constants and so they do not irritate us to get the derivation in this function. For the rest use @Nik's leading hint in the comment and this fact that $(UV)'=U'V+UV'$. Personally, I like another aswer because the chain rule makes your function simpler in form.

Nice work, and great sportsmanship, as always! – amWhy Sep 25 '13 at 00:05 — amWhy, Sep 25 '13 at 00:05

score 0 · Answer 2 · answered Sep 24 '13 at 06:58

Hint: Using chain rule, $\frac{d}{dx} \exp(-x^2)=\exp(-x^2)\cdot (-2x)$. Can you finish the problem now?

By the way the word derive has a different mathematical meaning than what you're referring to here. You should say, 'find the derivative of' or 'differentiate' in this context. Just saying.

Derivative of $\frac{d}{dx}\{\exp \frac{-(x-\mu)^2)}{2 \sigma^2}\}$

2 Answers2