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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.

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

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  • $\begingroup$ Hint: Chain rule $\endgroup$ – Jakobian Sep 27 '18 at 20:35
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Assuming $u = \sigma^2$, then:

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

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