# Derive vega for Black-Scholes call from this formula?

Is it possible to get the right formula for vega of a call option under the black scholes model from this formula?

$$\frac{\partial{C}}{\partial{\sigma}}=\frac{S_0}{\sqrt{2\pi}}{e^\frac{-d_+^2}{2}}(\frac{-1}{\sigma})(d_-)-\frac{Ke^{-rt}}{\sqrt{2\pi}}e^{\frac{-d_-^2}{2}}(\frac{-1}{\sigma})(d_+)$$

$d_-=\frac{\ln{\frac{S_0}{k}}+(r-\frac{\sigma^2}{2})t}{\sigma\sqrt{t}}$ $d_+=\frac{\ln{\frac{S_0}{k}}+(r+\frac{\sigma^2}{2})t}{\sigma\sqrt{t}}$

The answer by @Gordon is pretty complete, but let me add one more point. Let $$n(x) = N'(x)$$ be the PDF of standard normal distribution.
In the derivation, note that $$e^{d_+^2/2 - d_-^2/2} = \frac{n(d_-)}{n(d_+)} = \frac{S_0}{Ke^{-rt}}.$$ Thanks to this relation, there are two equivalent expressions for the Black-Scholes vega: $$\frac{\partial C}{\partial \sigma} = S_0 n(d_+) \sqrt{t} = K e^{-rt} n(d_-) \sqrt{t}.$$ See Wikipedia.