F(x) = -1/cos5x ;
f '(x) =( -1/cos5x) ' = -(1/cos5x) ' = - ((cos5x)^(-1) ) ' =
-(-1)*(cos5x)^(-2) *(cos5x)' =1/(cos5x)² *(-sin5x)*(5x)' =
-5sin5x/cos²5x = -5tq5x/cos5x.
--- или
f '(x) =( -1/cos5x) ' =( (-1)'*cos5x -(-1)*(cos5x)' )/(cos5x)² =
=(0 + (-sin5x)*(5x)')/cos²5x = -5sin5x/cos²5x = -5tq5x/cos5x.