# Simplifying the Derivative of a European Call Option

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.

• You are switching notation for strike between $X$ and $K$. I use $X$ in my answer. – RRL Feb 14 at 6:54
• Also $q$ and $d$ are both being used above to represent the dividend yield in the Black-Scholes formula. – RRL Feb 14 at 7:23
• @RRL I apologize for the lack of consistency and appreciate the assistance! – strwars Feb 14 at 12:18
• I believe I have edited out the inconsistencies – strwars Feb 14 at 12:27
• You're welcome. – RRL Feb 14 at 16:46

## 1 Answer

Since $$d_1 = d_2 + \sigma \sqrt{T}$$, we have

$$N’(d_1) = \frac{1}{\sqrt{2\pi}}e^{-d_1^2/2} \\ = \frac{1}{\sqrt{2\pi}}e^{-d_2^2/2}e^{-d_2\sigma\sqrt{T}}e^{-\sigma^2T/2} \\ = N’(d_2)\frac{X}{S}e^{-rT}e^{dT}$$

Substituting for $$N'(d_1)$$ in the last formula before “Finding the partial derivatives ...” and using

$$\frac{\partial d_1}{\partial X} - \frac{\partial d_2}{\partial X} = \sigma’(X)\sqrt{T}$$

will give you the answer without any further differentiation.