# Differentiating the risk-neutral price of a European call

(Black-Scholes formula) The risk-neutral price of a European call is $$C_t = S_tN(d_1) - e^{r\tau}KN(d_2)$$ where $$d_1 = \frac{log(\frac{S_t}{K}) + (r + \frac{1}{2}\sigma^2)}{\sigma\sqrt{\tau}}$$ and $$d_2 = d_1 - \sigma \sqrt{\tau}$$

$$N(x)$$ is the cdf of the normal distribution. I would like to differentiate $$C_t$$ with respect to $$S_t$$, the stock price at time t, to obtain $$\frac{\delta C_t}{\delta S_t} = N(d_1)$$

I've done $$C_t = S_tN(d_1) - e^{r\tau}KN(d_1 - \sigma \sqrt{\tau})$$ then $$\frac{\delta C_t}{\delta S_t} = S_td_1N(d_1) + N(d_1) - e^{-r\tau}Kd_1N(d_1)$$

but then does $$S_t = e^{-r\tau}K$$ to get the required result?

K is the strike price, $$r$$ is the interest rate and $$\tau = T - t$$, where $$T$$ is the end of the contract.

Any help would be appreciated thanks.

$$\phi$$ is the normal pdf, $$\Phi$$ is the normal cdf. We get $$\frac{\partial C}{\partial S}=S\frac{\partial d_1}{\partial S}\phi(d_1)+\Phi(d_1)-Ke^{-r(T-t)}\frac{\partial d_2}{\partial S}\phi(d_2)$$ We notice that $$\frac{\partial d_1}{\partial S}=\frac{\partial d_2}{\partial S}$$ (follows from the definition) so $$\frac{\partial C}{\partial S}=\Phi(d_1)+\frac{\partial d_1}{\partial S}\bigg(S\phi(d_1)-Ke^{-r(T-t)}\phi(d_2)\bigg)$$ It suffices to show that the term in the parenthesis is $$0$$. We get \begin{aligned} S\phi(d_1)&=\frac{1}{\sqrt{2\pi}}\exp\bigg(\ln(S)+\frac{-(\ln(S/K)+(r+\sigma^2/2)(T-t))^2}{2\sigma^2(T-t)}\bigg)=\\ &=\frac{1}{\sqrt{2\pi}}\exp\bigg(\ln(S)+\frac{-(\ln(S/K)+(r-\sigma^2/2)(T-t)+\sigma^2(T-t))^2}{2\sigma^2(T-t)}\bigg)=\\ &=\frac{1}{\sqrt{2\pi}}\exp\bigg(\ln(S)-\frac{(\ln(S/K)+(r-\sigma^2/2)(T-t))^2}{2\sigma^2(T-t)}-\frac{\sigma^2(T-t)}{2}+\\ &-(\ln(S/K)+(r-\sigma^2/2)(T-t))\bigg)=\\ &=\frac{1}{\sqrt{2\pi}}\exp\bigg(\ln(K)-r(T-t)-\frac{(\ln(S/K)+(r-\sigma^2/2)(T-t))^2}{2\sigma^2(T-t)}\bigg)=\\ &=Ke^{-r(T-t)}\phi(d_2) \end{aligned}