I have a task to do but it is very difficult.. I have to calculate the:
How can I obtain that?
I'm not sure I agree with that being a very difficult task...
The black formula for a caplet (using notation from Hull's book) is given by:
$caplet = L \delta_k P(0, t_{k+1}) [F_k N(d_1) - R_kN(d_2)]$
where:
$d_1 = \frac{ln(F_k/R_k) + \sigma_k^2t_k/2}{\sigma_k\sqrt{t_k}}$ and $d_2 = d_1 - \sigma_K \sqrt{t_k}$
The delta will just be the first derivative of the equation above in respect to F, ie:
Delta = $ L \delta_k P(0, t_{k+1}) N(d_1)$
You can test this by comparing the calculated Delta * 1 basis point on one side with the caplet price for forward + 1 basis point minus the caplet price for forward - 1 basis point divided by two.
For the vega, just do the same thing, but derive in respect to $\sigma$. (hint: the sigma is in the $d_1$)
Edit after comments:
As Jan Stuller very well pointed out, the derivation of the well known result for the delta is, in fact, not as simple as it seems, although not that difficult.
Ignoring the constants $L \delta_k P(0, t_{k+1})$ ...
$$ \Delta = \frac{\partial c}{\partial F} = N(d_1) + F \frac{\partial N(d_1)}{\partial F} - R_k \frac{\partial N(d_2)}{\partial F}$$
Next we apply the chain rule:
$$ \frac{\partial N(d_1)}{\partial F} = \frac{\partial N(d_1)}{\partial d_1} \frac{\partial d_1}{\partial F}$$
and so:
$$ \Delta = N(d_1) + F \frac{\partial N(d_1)}{\partial d_1} \frac{\partial d_1}{\partial F} - R_k \frac{\partial N(d_2)}{\partial d_2} \frac{\partial d_2}{\partial F}$$
First thing to notice is that since:
$d_1 = \frac{ln(F_k/R_k) + \sigma_k^2t_k/2}{\sigma_k\sqrt{t_k}}$ and $d_2 = d_1 - \sigma_K \sqrt{t_k}$
we have:
$$\frac{\partial d_1}{\partial F} = \frac{\partial d_2}{\partial F} = \frac{1}{F_k \sigma \sqrt{t_k}}$$
and so:
$$ \Delta = N(d_1) + \frac{1}{F_k \sigma \sqrt{t_k}} \left[ F \frac{\partial N(d_1)}{\partial d_1} - R_k \frac{\partial N(d_2)}{\partial d_2} \right] $$
Second, lets see how we can simplify:
$$ \left[ F \frac{\partial N(d_1)}{\partial d_1} - R_k \frac{\partial N(d_2)}{\partial d_2} \right]$$
Considering that:
$$N(d_1) = \frac{1}{\sqrt{2\pi}} \int^{d_1}_{-\infty} e^{-x^2 / 2} dx$$
it follows that:
$$\frac{\partial N(d_1)}{\partial d_1} = N^{'}(d_1) = \frac{1}{\sqrt{2\pi}} e^{-d_1^2 / 2}$$
and so:
$$ \left[ F \frac{\partial N(d_1)}{\partial d_1} - R_k \frac{\partial N(d_2)}{\partial d_2} \right] = \left[ FN^{'}(d_1) - R_k N^{'}(d_2) \right]$$
Now with a simple substitution of $d_2$ for $d_1 - \sigma_k \sqrt{t_k}$, we have:
$$R_k N^{'}(d_2) = R_k N^{'}(d_1 - \sigma_k \sqrt{t_k})$$
$$ = R_k \frac{1}{\sqrt{2\pi}} e^{\frac{-(d_1 - \sigma_k \sqrt{t_k})^2}{2}} = R_k \frac{1}{\sqrt{2\pi}} e^{\frac{-d_1^2}{2}} e^{ \frac{-(2d_1 \sigma_k \sqrt{t_k} + \sigma_t^2 t_k)}{2}}$$
$$ = R_k N^{'}(d_1) e^{ \frac{ln(F_k/R_k) + \sigma_k^2t_k/2}{\sigma_k\sqrt{t_k}} \sigma_k \sqrt{t_k}} e^{- \sigma_t^2 t_k / 2)} = R_k N^{'}(d_1) e^{ ln(F_k/R_k) + \sigma_k^2t_k/2} e^{- \sigma_t^2 t_k / 2)}$$
$$ = R_k N^{'}(d_1) e^{ ln(F_k/R_k) + \sigma_k^2t_k/2 - \sigma_t^2 t_k / 2)} = R_k N^{'}(d_1) \frac{F_k}{R_k} = F_k N^{'}(d_1)$$
$$ $$
and so
$$R_k N^{'}(d_2) = F_k N^{'}(d_1)$$
$$F_k N^{'}(d_1) - R_k N^{'}(d_2) = 0$$
So finally, the delta of the caplet would be:
$$ \Delta = N(d_1) + \frac{1}{F_k \sigma \sqrt{t_k}} \left[ FN^{'}(d_1) - R_k N^{'}(d_2) \right] = N(d_1) + \frac{1}{F_k \sigma \sqrt{t_k}} \times 0 $$
$$ \Delta = N(d_1)$$
