Delta=N(d1) ???

[quotes]

I found the definition of delta with respect to an European option problematic:

(\frac{\partial C}{\partial S_t}=\frac{\partial}{\partial S_t}(S_t N(d1)-Ke^{-r(T-t)}N(d2))=N(d1))

while d1 and d2 are actually functions of S.

Shall we count in the derivatives of N(d1) and N(d2) ?

Or the partial derivative approach is just a heuristic explanation for Martingle Representation Theorem?

 

[quotes]

If the derivatives of N(d1) or N(d2) are not in consideration,

the partial derivative on the latter, negation of a share digital option is zero as well, which contradicts reality.

 

[MrBucks]

I found the definition of delta with respect to an European option problematic:

(\frac{\partial C}{\partial S_t}=\frac{\partial}{\partial S_t}(S_t N(d1)-Ke^{-r(T-t)}N(d2))=N(d1))

while d1 and d2 are actually functions of S.

Shall we count in the derivatives of N(d1) and N(d2) ?

Or the partial derivative approach is just a heuristic explanation for Martingle Representation Theorem?

(\frac{\partial C}{\partial S} = N \left ( d_{1} \right ) + S \frac{\partial N \left ( d_{1} \right )}{\partial d_{1}} \frac{\partial d_{1}}{\partial S} - Ke^{-r \left ( T - t \right )} \frac{\partial N \left ( d_{2} \right )}{\partial d_{2}} \frac{\partial d_{2}}{\partial d_{1}} \frac{\partial d_{1}}{\partial S})

When you do the algebraic manipulations at this point, you will find that everything after the (N \left ( d_{1} \right )) equals zero.

 

[quotes]

When you do the algebraic manipulations at this point, you will find that everything after the (N \left ( d_{1} \right )) equals zero.

Oh thank you Bucks, I just remembered my professor did mention this problem and I found in John Hull 6e Homework 13.17 on this matter. Yes they cancel out.