This is a little trickier and requires the use of negation elimination, which is properly calleddouble-negation elimination. It is easier to derive (¬p ∨ ¬q) if you can derive (A ∨¬A) for any formula A. But assuming this is not done, we proceed as follows.

Assume ¬(¬p ∨ ¬q). From this in the end we will derive a contradiction and thus ¬¬(¬p ∨ ¬q), and then use double-negation elimination to derive (¬p ∨ ¬q). Now assume p and q. Then p ∧ q, which contradicts ¬(p ∧ q). Therefore, ¬p, which implies (¬p ∨ ¬q). This in turn contradicts ¬(¬p ∨ ¬q), so we conclude ¬q. Then we repeat by deriving (¬p ∨ ¬q), which again contradicts ¬(¬p ∨ ¬q). Thus, ¬¬(¬p ∨ ¬q) and (¬p ∨ ¬q).