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.
Assuming $u = \sigma^2$, then:
$$\frac{dy}{d\sigma}=\frac{dy}{du} \cdot\frac{du}{d\sigma} =\frac{dy}{du}\cdot (2\sigma)$$ So $$\frac{dy}{du}=\frac{1}{2\sigma} \cdot \frac{dy}{d\sigma}$$