The BSM formula for a call is

\[\begin{aligned}C(0) & =e^{-r T}\left[F_{A}(0, T) N\left(d_{1}ight)-K N\left(d_{2}ight)ight] \\& =A(0) N\left(d_{1}ight)-K e^{-r T} N\left(d_{2}ight)\end{aligned}\]

One might be tempted to calculate \(N\left(d_{1}ight)\) as the delta, i.e.

\[\frac{\partial C}{\partial A}=N\left(d_{1}ight)\]

However, this overlooks the fact that \(d_{1,2}\) are functions of \(F_{A}(0, T)=\) \(A(0) e^{r T}\). Calculate delta using the chain rule as follows. Let \(F=F_{A}(0, T)\) for notation convenience. Since \(C(0) e^{r T}=F N\left(d_{1}ight)-K N\left(d_{2}ight)\) and \(F=A(0) e^{r T}\)
\[\begin{aligned}\frac{\partial C}{\partial A} & =\frac{\partial\left[F N\left(d_{1}ight)-K N\left(d_{2}ight)ight]}{\partial F} \\& =N\left(d_{1}ight)+F N^{\prime}\left(d_{1}ight) \frac{\partial d_{1}}{\partial F}-K N^{\prime}\left(d_{2}ight) \frac{\partial d_{2}}{\partial F}
\end{aligned}\]

(a) Compute the terms \[\frac{\partial d_{1}}{\partial F}, \quad \frac{\partial d_{2}}{\partial F}\]

(b) Complete the following steps and solve for \(Z\)
\[\begin{aligned}N^{\prime}\left(d_{1,2}ight) & =\frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2} d_{1,2}^{2}} \\& \ldots \\& =\frac{1}{\sqrt{2 \pi}}\left(\frac{K}{F}ight)^{ \pm 1 / 2} Z \end{aligned}\]

(c) Using the above results, provide an expression for \(\partial C / \partial A\).