Differentiate the product of some terms

As stated in my comment, the hard thing here is, that Mathematica cannot expand your Product for an unknown k. What you know is, that $1\leq j \leq k$ but I don't how I could help Mathematica, that it expands your Product with this knowledge. What we want to do is the following:

$$\prod_{i=1}^k (1+\lambda_i)e^{\lambda_i} \rightarrow \\\left(\prod_{i=1}^{j-1} (1+\lambda_i)e^{\lambda_i}\right) \cdot \left((1+\lambda_j)e^{\lambda_j}\right)\cdot \left(\prod_{i=j+1}^k (1+\lambda_i)e^{\lambda_i}\right)$$

Let's try to do this small part manually

expr = Product[(1 + l[i]) Exp[l[i]], {i, k}];
exprExpanded = expr /. Product[f_, {i_, k_}] :> 
  Product[f, {i, 1, j - 1}]*(f /. i :> j)*Product[f, {i, j + 1, k}]

$$e^{l[j]} (1+l[j]) \left(\prod _{i=1}^{-1+j} e^{l[i]} (1+l[i])\right) \prod _{i=1+j}^k e^{l[i]} (1+l[i])$$

In this form we can derive the expression with D because everything which depends on l[j] is extracted

dexpr = D[exprExpanded, l[j]]

$$e^{l[j]} \left(\prod _{i=1}^{-1+j} e^{l[i]} (1+l[i])\right) \prod _{i=1+j}^k e^{l[i]} (1+l[i])+e^{l[j]} (1+l[j]) \left(\prod _{i=1}^{-1+j} e^{l[i]} (1+l[i])\right) \prod _{i=1+j}^k e^{l[i]} (1+l[i])$$

This is the general analytic form of your derivative. Now you can check whether it is equivalent to calculated derivative when you insert special values for j and k

(dexpr /. {j :> 3, k :> 5}) == D[Product[(1 + l[i]) Exp[l[i]], {i, 5}], l[3]]
(*
 True
*)