# derive black scholes greeks

I am reading a paper and get a problem here, the following terms are all from standard BS models. the paper says using the well known fact $$Se^{-q(T-t)}N^{'}(d1)=Ke^{-r(T-t)}N^{'}(d2)$$ here the differentiation is respect to $T$.For instance $N^{'}(T^2+1)=2T$

So anyone could give some hints to get this fact?

• I suppose $q$ is a continuous yield dividend and $r$ the risk-free rate. But be careful, $N^\prime$ is most likely not the differential operator here! It is the probability density function of the standard normal. Apr 25, 2014 at 12:33

The first equality is a bit tedious to derive but straight-forward. As commented by vanguard2k, the notation $N'(x)$ is meant to denote the density function of the standard normal distribution: $$N'(x) = n(x) := \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$$. Simply insert the $d_{i}$ terms, $$d_1 = \frac{\log\left(\frac{S}{K}\right) + (r-q+\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}$$ $$d_2 = \frac{\log\left(\frac{S}{K}\right) + (r-q-\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}}$$ and rearrange.