Deriving Black–Scholes Call Theta by Differentiating the Pricing Formula

Summary

The document asks how to derive the time derivative, or theta, of a European call from the Black–Scholes pricing formula. It provides the call price in terms of the normal cumulative distribution function and the expressions for the two standardized variables, then states the resulting theta formula. The questioner understands the derivative of the discounted-strike term but is unsure how differentiating the variable involving time to maturity produces the volatility-and-spot-price component.

This is a focused calculus question rather than a worked derivation: it gives the starting equation and target expression but no answer, intermediate steps, numerical example, or independent evidence. Readers can identify the chain-rule issue that needs resolving, but the document alone does not explain the cancellation that simplifies the derivative. The formula also uses a particular convention in which time is calendar time and maturity is fixed, so theta is expressed as a derivative with respect to valuation time.

Key ideas

• Call theta is obtained by differentiating the Black–Scholes call price with respect to valuation time.
• The time-to-maturity terms in both standardized variables must be differentiated using the chain rule.
• The stated theta contains a discounted-strike contribution and a volatility contribution involving the normal density.
• The document poses the derivation question but does not provide its solution.

Tags

# How to derive the Greek theta from Black-Scholes solution formula?


# How to derive the Greek theta from Black-Scholes solution formula?












Which are the steps to compute the theta greek from the BS solution:

$$c(t, x) = xN(d_+(T-t,x)) - K e ^{-r(T-t)}N(d_-(T-t,x))$$

with:

$$ d_\pm (T-t, x) = \dfrac{1}{\sigma \sqrt{T-t}} \left[ \ln \left( \dfrac{x}{K} \right) + \left( r \pm \dfrac{\sigma^2}{2} \right) (T-t) \right] $$

I know that the answer is:

$$ c_t(t,x) = -rKe^{-r(T-t)}N(d_-(T-t,x)) - \dfrac{\sigma x}{2 \sqrt{T-t}}N'(d_+(T-t,x)) $$

Now, form me it is clear how to obtain the first term: $-rKe^{-r(T-t)}N(d_-(T-t,x))$; the problem is how I can derive $d_-(T-t,x)$ in order to obtain:

$$ - \dfrac{\sigma x}{2 \sqrt{T-t}} $$

Thanks in advance.

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