# Derivative of a Standard Normal CDF [closed]

I have a quite simple question to ask. Consider a function $$y(\sigma)=\ln\Phi(-\frac{c}{\sigma})$$, where the function $$\Phi(\cdot)$$ is the standard normal CDF, $$c$$ is a constant number and $$\sigma\geq 0$$ is a random variable.

Questions: what is the derivative of $$\frac{\partial y(\sigma)}{\partial \sigma^2}=?$$

I got confused when taking the derivative with respect to $$\sigma^2$$, not $$\sigma$$. Thank you in advance.

## closed as off-topic by Did, Adrian Keister, Namaste, Deepesh Meena, Xander HendersonSep 28 '18 at 3:02

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• Hint: Chain rule – Jakobian Sep 27 '18 at 20:35

## 1 Answer

Assuming $$u = \sigma^2$$, then:

$$\frac{dy}{d\sigma}=\frac{dy}{du} \cdot\frac{du}{d\sigma} =\frac{dy}{du}\cdot (2\sigma)$$ So $$\frac{dy}{du}=\frac{1}{2\sigma} \cdot \frac{dy}{d\sigma}$$