With the Black-Scholes formula for a call option:
$$P= S e^{-\delta T } \Phi(d_{1}) - K e^{-rT} \Phi(d_{2}) $$
With $d_{1}$ and $d_{2}$ as:
$$ d1 = \frac{\ln\left(\frac{S}{K}\right) + \left(r - \delta + \frac{\sigma^2}{2}\right)T}{\sigma \sqrt{T}} $$
$$ d2 = d1 - \sigma \sqrt{T} $$
I derived $\frac{\partial P}{\partial S}$ and ended up with this:
$$ \Delta = e^{-\delta T} \Phi(d_{1}) + \frac{S e^{-\delta T} \phi(d_{1})- K e^{-r T} \phi(d_{2}) }{S \sigma \sqrt{T}} $$
I know the formula for delta is: $\Delta= e^{-\delta T} \Phi(d_{1})$ so that means the right side of the sum equals zero. Also in the Book: "Derivatives Markets - Mc Donald" says it can be shown that:
$$ S e^{-\delta T} \phi(d_{1})= K e^{-r T} \phi(d_{2}) $$
This is where i cant follow up, i dont see how they are equal and i havent found any book with this proof. I know its equal, but why?
Too long for a comment.
Just complete the square. You know that $$ \phi(x) = \frac{1}{\sqrt{2\pi}} \, \exp\left\{ -\frac12 x^2\right\} $$ and $$ d_1 = \frac{ \log S_t/K + (r - \delta + \frac12 \sigma^2)T }{ \sigma\sqrt T}, \quad d_2 = d_1 - \sigma \sqrt T $$ It follows then quite easily that $$ Se^{-\delta T} \phi (d_1) = Ke^{-r T} \phi (d_2) $$