# Derivative of a std Normal CDF?

I was wondering how I can find the derivative of a normal cdf with respect to a boundary parameter?
I can get an answer with Mathematica or something but I have no idea how to actually do this. I don't know how fundamental theorem of calculus can be applied. (if it can be)

$\Phi$((log(S/K)+(r+σ^2)T)/(σ sqrt(T)))

where phi is the std normal cdf.

I'm doing option pricing analysis but I can't really figure out how to derive this with respect to r or sigma or any other parameter. I would appreciate any help.

mathman
The standard normal distribution is usually expressed in terms of 2 parameters, the mean and variance. Your expression has one parameter, so I can't tell what you mean.

standard normal has a mean 0 and variance 1.

So basically I need to derive
∫$\frac{1}{\sqrt{2\pi}}$exp($\frac{-x^2}{2}$) dx

with upper bound
(log(S/K)+(r+σ^2)T)/(σ sqrt(T))
lower bound -inf

with respect to r, Sigma or any parameter so I can learn how to do this.

mathman
The cdf of the normal distribution cannot be expressed analytically. To use it you need to calculate the upper bound explicitly and get the answer from a table.

How do you think those values were found?
The derivative with respect to r is this... according to Mathematica. But I don't know how.

$\frac{\sqrt{T}}{\sqrt{2\pi}σ}$exp(-($\frac{(T (r + σ^2) + Log[S/K] )}{\sqrt{2 T σ}}$)^2)

with changing a few things ==

T * Normal PDF(-log(S/K),Tσ^2) at point T*(r+^2)

Technically this is suppose to be 0 as normal pdf is 0 at any point since it is continuous but something different can be acquired with deriving with respect to something else.

Mute
Homework Helper
Just use the chain rule. e.g., for r

$$\frac{d\Phi(f(r))}{dr} = \frac{d\Phi(y)}{dy} \frac{dy(r)}{dr},$$

where y = f(r) is the argument of your cdf. Since it's a standard normal distribution, $d\Phi(y)/dy = \exp(-y^2/2)/\sqrt{2\pi}$, and then plug in y = f(r), of course.

You can do a similar thing treating the argument of the cdf as a function of $\sigma$.

Thank you. That was of great help!