How to differentiate the standard normal deviation w.r.t. a parameter inside the upper bound

Question

Given that $$N(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{s^2}{2}}\:ds$$

And that $$d=\frac{1}{\sigma\sqrt{\tau}}\ln\left({\frac{S}{e^{-r\tau}K}}\right)+\sigma\sqrt{\tau}$$

How do I take the derivative of $N(d)$ with respect to $\sigma$?

$$\frac{dN(d)}{d\sigma}=?$$

Answer 1

We have $$N(f(\sigma)) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{f(\sigma)}exp \left(-\frac{s^2}{2}\right)ds.$$ Thus, using the chain rule:

$$\frac{\partial N(f(\sigma))}{\partial \sigma} = \frac{\partial N(f(\sigma))}{\partial f(\sigma)}.\frac{\partial f(\sigma)}{\partial \sigma} = \frac{1}{\sqrt{2\pi}}exp\left(-\frac{f(\sigma)^2}{2}\right)\left[-\frac{1}{\sigma^2 \sqrt{\tau}}ln\left(\frac{S}{exp\left(-rt\right)K}\right)+\sqrt{t}\right],$$ where $$f(\sigma) =\frac{1}{\sigma \sqrt{t}}ln\left(\frac{S}{exp\left(-rt\right)K}\right)+\sigma\sqrt{t}.$$

Notation: $exp{(\cdot)} = e^{(\cdot)}$.