# Derivation of stock price formula John C. Hull 9th Ed p309

It says assuming a no-uncertainty Weiner process that models stock price: $$\Delta S = \mu S\Delta t$$ Can be rearranged to (after taking the limit of $$\Delta t \to 0$$... $$\frac{dS}{S}=\mu dt$$ Then integrating between time 0 and T to get: $$S_T=S_0 e^{\mu T}$$

I don't understand the last step. Are they integrating with respect to t? How does the exponential come about when there was no exponential in the prior equation? Is this step a condensation of a complex calculation that they didn't show?

You can naively put integral signs on both sides of the equation: \begin{align*} \frac{\mathrm{d}S_t}{S_t} &=\mu \mathrm{d}t \\ \implies \int_0^T\frac{\mathrm{d}S_t}{S_t} &=\int_0^T\mu \mathrm{d}t\\ \implies \ln(S_T)-\ln(S_0) &=\mu T \\ \implies S_T&=S_0e^{\mu T}. \end{align*}
Perhaps this makes it easier: Since $$S_t$$ is deterministic, it is a normal'' function. Thus, you may want to write $$y(x)=S_t$$. The above equation then turns into $$\frac{\mathrm{d}y}{y} =\mu \mathrm{d}x\Leftrightarrow y'=\mu y$$. So, it's simply about solving a first-order ODE.
Also note that $$\frac{\mathrm{d}S_t}{S_t}=\mathrm{d}\ln(S_t)$$, i.e. percentage returns correspond to log-returns if time is infinitesimal. So, you shouldn't be surprised to find an exponential here.