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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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    $\begingroup$ Differentiate w.r.t. $t$ and carefully apply the chain rule. Where do you get stuck when you try this? $\endgroup$ – LocalVolatility Aug 30 '17 at 10:45
  • $\begingroup$ I am not able to conclude the derivation in the correct way. $\endgroup$ – Archimede Aug 30 '17 at 12:33
  • $\begingroup$ Then start by showing us how far you got. It is much easier to point out where you went wrong (especially if it is somewhere near the end) than typing up everything from scratch. $\endgroup$ – LocalVolatility Aug 30 '17 at 12:34
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    $\begingroup$ 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!). $\endgroup$ – LocalVolatility Aug 30 '17 at 12:55
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    $\begingroup$ See also answers here and here. $\endgroup$ – Gordon Aug 30 '17 at 14:23

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