# Verifying an identity of an equation for Black Scholes formula

I just started working on the Black Scholes formula with help of the book Financial option valuation by Higham. Apparently you are possible to derive the following function:

$\log(\frac{SN'(d_1)}{e^{-r(T-t)}EN'(d_2)}) = 0$

From the Black scholes formula:
$C(S,t)=SN(d_1)-Ee^{-r(T-t)}N(d_2)$

I've been puzzling arround but I'm stuck. This is where I came so far, do you know where I'm going wrong?

$\log(\frac{SN'(d_1)}{e^{-r(T-t)}EN'(d_2)}) = \log(SN'(d_1))-\log(e^{-r(T-t)}EN'(d_2))=0$

• $N'$ is the derivatve of the normal cdf - right? What if you plug in? – Richard Oct 22 '15 at 12:49
• Yes, $N'$ is the derivative of a normal cdf. So you could rewrite that as $N'(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$ But what do you mean with "plug in"? – Alfons Ingomar Oct 22 '15 at 13:55
• Keep going with the substitution, you will see. – noob2 Oct 22 '15 at 14:03
• You plug-in $d1$ and $d2$ ... – Richard Oct 22 '15 at 14:03
• Please don't cross post your question, see this topic for guidance. – Bob Jansen Oct 26 '15 at 21:21

The numerator is $$S N'(d_1) = S \frac{1}{\sqrt{2 \pi}} \exp(-1/2 d_1^2) = \\ S \frac{1}{\sqrt{2 \pi}} \exp\left(- \frac12 \left(\log(S/E)+ (r + \frac12 \sigma^2(T-t)) \right)^2 / \sigma^2 (T-t) \right)$$ the denominator is: $$\exp(-r (T-t)) E N'(d_2) = \\ E \frac{1}{\sqrt{2 \pi}} \exp\left(- \frac12 \left(\log(S/E)+(r- \frac12 \sigma^2(T-t)) \right)^2 / \sigma^2 (T-t) -r(T-t) \right).$$ Now what if we extend the square, match exp and log and if then nominator and denominator are equal we get the result.