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:

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

• Differentiate w.r.t. $t$ and carefully apply the chain rule. Where do you get stuck when you try this? – LocalVolatility Aug 30 '17 at 10:45
• Your mistake is that both $N(d_+)$ and $N(d_-)$ are functions of $t$ . Differentiate each of them separately and then simplify the expression you get. You need the identity $S_0 \mathcal{N}’ ( d_+ ) = K e^{-r (T - t)} \mathcal{N}’ ( d_- )$ along the way (prove it yourself!). – LocalVolatility Aug 30 '17 at 12:55