I am studying about Derivation of the Black–Scholes formula. I have two questions. First one: please check my writing if it's correct, and the second one: Is there another method to obtain the Black–Scholes formula? The price of the underlying asset follows a geometric Brownian motion. That is *$\frac{dS}{S}=\mu dt+\sigma dB_t$ where B is a Brownian motion.
Remark ** discrete form $\Delta S = \mu S \Delta t + \sigma S \Delta B_t $
The payoff of an option $V(S,T)$ at maturity is known.To find its value at an earlier time we need to know how $V$ evolves as a function of $S$ and $t$. By $It \hat{o}'s$ lemma we have
$$dV=\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial S}ds+\frac 12\frac{\partial^2 V}{\partial S^2}(dS)^2\\
WRT*|\to dV=\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial S}(S\mu dt+S\sigma dB_t)+\frac 12\frac{\partial^2 V}{\partial S^2}(S\mu dt+S\sigma dB_t)^2\\
dV=\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial S}(S\mu dt+S\sigma dB_t)+\frac 12\frac{\partial^2 V}{\partial S^2}(S^2\mu^2 (dt)^2+S^2\sigma^2 (dB_t)^2+2S^2\mu\sigma (dt.dB_t))\\
$$ as we know $$(dt)^2 \to 0\\(dt.dB_t)\to 0\\(dB_t)^2\to dt$$so $$
dV=\frac{\partial V}{\partial t}dt+\frac{\partial V}{\partial S}(S\mu dt+S\sigma dB_t)+\frac 12\frac{\partial^2 V}{\partial S^2}S^2\sigma^2 dt\\
dV=(\frac{\partial V}{\partial t}+\frac{\partial V}{\partial S}S\mu +\frac 12\frac{\partial^2 V}{\partial S^2}S^2\sigma^2)dt+S\sigma dB_t$$
consisting of being short one option and long ${\frac {\partial V}{\partial S}}$ shares at time $t$. The value of these holdings is $$\pi=-V+\frac{\partial V}{\partial S}S$$ in the inetrval $[t,t+\Delta t]$, the total profit or loss from changes in the values of the holdings is
$$\Delta\pi=-\Delta V+\frac{\partial V}{\partial S}\Delta S$$ if we discretize $dV$ formula, and use * in it
$$\Delta V=(\frac{\partial V}{\partial t}+\frac{\partial V}{\partial S}S\mu +\frac 12\frac{\partial^2 V}{\partial S^2}S^2\sigma^2)\Delta t+S\sigma \Delta B_t$$
and substitute them into the expression for $\Delta \pi$
$$\Delta\pi=-\Delta V+\frac{\partial V}{\partial S}\Delta S\\
\Delta\pi=-(\frac{\partial V}{\partial t}+\frac{\partial V}{\partial S}S\mu +\frac 12\frac{\partial^2 V}{\partial S^2}S^2\sigma^2)\Delta t+S\sigma \Delta B_t)+\frac{\partial V}{\partial S}( \mu S \Delta t + \sigma S \Delta B_t)\\
\Delta\pi=-(\frac{\partial V}{\partial t}+\frac 12\frac{\partial^2 V}{\partial S^2}S^2\sigma^2)\Delta t$$
Now assuming the risk-free rate of return is $r$ we must have over the time period $[t,t+\Delta t]$ is :$r\pi \Delta t =\Delta \pi$
so $$r\pi \Delta t =\Delta \pi\\r(-V+\frac{\partial V}{\partial S}S)\Delta t=-(\frac{\partial V}{\partial t}+\frac 12\frac{\partial^2 V}{\partial S^2}S^2\sigma^2)\Delta t\\\div \Delta t \\
\frac{\partial V}{\partial t}+\frac12 \sigma^2 S^2 \frac{\partial^2 V}{\partial s^2}+rS\frac{\partial V}{\partial S}-rV=0$$ Thank you for reading my long writing. I am also looking for Alternative derivation method to obtain the formula, as far as I searched (6.4 of Shreve ) sugest by using Ito's formula on $\exp(-rt)V(t,S(t))$, But honesly I stuck at this approach ( I don't how to use it ).Please help me at this or guide me for an alternative method.
Thanks in advance.