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You know $$N(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{\frac{-u^2}{2}}du$$ then $$N'(x)=\frac{1}{\sqrt{2\pi}}e^{\frac{-u^2}{2}}$$ therefore $$\mathcal{V}=S_t\sqrt{\tau}N'(d_1)$$



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