I know the title suggests finance, but I'm stuck on the mathematics of this.
I need to take the following derivative:
$$ -\frac{\delta C(X)}{\delta X}=-\frac{\delta}{\delta X} \Big[Se^{-dT}N(d_1)-Xe^{-rT}N(d_2)\Big] $$
where S, d, and T are constants,
$$ d_1 = d_2 + \sigma \sqrt{T} = \frac{ln(\frac{S}{X})+(r-d+\frac{\sigma^2}{2})T}{\sigma \sqrt{T}} $$
$$ d_2 = \frac{ln(\frac{S}{X})+(r-d-\frac{\sigma^2}{2})T}{\sigma \sqrt{T}} $$
and $\sigma$ is a function of X such that $\sigma = \sigma(X)$. Further, I need to show this derivative ultimately equals
$$ -\frac{\delta C(X)}{\delta X} = e^{-rT}N(d_2) - Xe^{-rT}N'(d_2) \sqrt{T} \sigma'(X) $$
I have currently done the following
$$ \begin{align} -\frac{\delta C(X)}{\delta X} &=-\frac{\delta}{\delta X} \Big[Se^{-dT}N(d_1)-Xe^{-rT}N(d_2)\Big] \\ &=-\frac{\delta}{\delta X} \Big[Se^{-dT}N(d_1)\Big]+\frac{\delta}{\delta X}\Big[Xe^{-rT}N(d_2)\Big] \\ &= -Se^{-dT}\frac{\delta}{\delta X} \Big[N(d_1)\Big]+e^{-rT}\frac{\delta}{\delta X}\Big[XN(d_2)\Big] \\ &= -Se^{-dT}\frac{\delta N(d_1)}{\delta d_1}\frac{\delta d_1}{\delta X} + Xe^{-rT}\frac{\delta N(d_2)}{\delta d_2}\frac{\delta d_2}{\delta X} + e^{-rT}N(d_2) \end{align} $$
Finding the partial derivatives
$$ \frac{\delta N(d_1)}{\delta d_1} = N'(d_1) $$
$$ \frac{\delta N(d_2)}{\delta d_2} = N'(d_2) $$
$$ \frac{\delta d_1}{\delta X} = \frac{\delta d_2}{\delta X} + \frac{\delta}{\delta X}\Big[\sigma(X) \sqrt{T} \Big] = \frac{\delta d_2}{\delta X} + \sigma'(X) \sqrt{T} $$
$$ \begin{align} \frac{\delta d_2}{\delta X} &= \frac{\delta}{\delta X} \Bigg[ \frac{ln(\frac{S}{X})+(r-d-\frac{\sigma(X)^2}{2})T}{\sigma(X) \sqrt{T}} \Bigg] \\ &= \frac{\delta}{\delta X}\Bigg[\frac{ln(\frac{S}{X})}{\sigma(X) \sqrt{T}}\Bigg] + \frac{\delta}{\delta X}\Bigg[\frac{(r-d)T}{\sigma(X)\sqrt{T}}\Bigg] + \frac{\delta}{\delta X}\Bigg[\frac{\frac{\sigma(X)^2T}{2}}{\sigma(X)\sqrt{T}}\Bigg] \\ &= \frac{ \frac{\sigma(X)}{X}-\sigma'(X)ln(\frac{S}{X})}{\sigma(X)^2 \sqrt{T}} + (r-d)\sqrt{T} \frac{\sigma'(X)}{\sigma(X)^2} + \frac{\sigma'(X) \sqrt{T}}{2} \end{align} $$
Applying these partial derivatives to the derivation
$$ \begin{align} -\frac{\delta C(X)}{\delta X} &= -Se^{dT}N'(d_1)\Bigg[ \frac{\delta d_2}{\delta X} + \sigma'(X) \sqrt{T} \Bigg] + Xe^{-rT}N'(d_2) \frac{\delta d_2}{\delta x} + e^{-rT}N(d_2) \\ &= -Se^{-dT}\Bigg[N(d_1)\frac{\delta d_2}{\delta X} + \sigma'(X) \sqrt{T} \Bigg] + Xe^{-rT}N'(d_2)\frac{\delta d_2}{\delta X} + e^{-rT}N(d_2) \end{align} $$
From here it appears that I need the first term to zero out and $\frac{\delta d_2}{\delta X} = \sqrt{T} \sigma'(X)$, but after working with the formulae for a while I've gotten no closer to an answer.