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The intermediate variable $d_1$ is calculated using the formula $d_1 = \frac{\ln(S_0/X) + (r + \sigma^2/2)T}{\sigma\sqrt{T}}$. The intermediate variable $d_2$ is not computed from scratch; instead, it is derived directly from $d_1$ using the relationship $d_2 = d_1 - \sigma\sqrt{T}$.

Conditions: $S_0 > 0$ and $X > 0$ to evaluate the natural logarithm $\ln(S_0/X)$.; $\sigma > 0$ to serve as a valid denominator.; $T > 0$ for the square root $\sqrt{T}$ to be meaningful.

The intermediate variable $d_1$ is calculated using the formula $d_1 = \frac{\ln(S_0/X) + (r + \sigma^2/2)T}{\sigma\sqrt{T}}$. The intermediate variable $d_2$ is not computed from scratch; instead, it is derived directly from $d_1$ using the relationship $d_2 = d_1 - \sigma\sqrt{T}$.

Conditions: $S_0 > 0$ and $X > 0$ to evaluate the natural logarithm $\ln(S_0/X)$.; $\sigma > 0$ to serve as a valid denominator.; $T > 0$ for the square root $\sqrt{T}$ to be meaningful.

The put option formula can be memorized by comparing it structurally with the call option formula. Starting from the call formula $C = S_0 N(d_1) - X e^{-rT} N(d_2)$, you swap the positions of the two terms and negate the arguments inside the standard normal distribution function $N(\cdot)$.

Conditions: You already know the Black-Scholes-Merton formula for the call option.; This is a mnemonic rule for memorization, not a rigorous mathematical proof.

The put option formula can be memorized by comparing it structurally with the call option formula. Starting from the call formula $C = S_0 N(d_1) - X e^{-rT} N(d_2)$, you swap the positions of the two terms and negate the arguments inside the standard normal distribution function $N(\cdot)$.

Conditions: You already know the Black-Scholes-Merton formula for the call option.; This is a mnemonic rule for memorization, not a rigorous mathematical proof.