The first is true, for obvious reasons.

(f + g)' = f' + g'

If f(x)+g(x) is differentiable at c then f and g must be defined and continuous at c, and therefore, each is independently differentiable at c.

The second may not be as easy to prove:

If f(x)*g(x) is differentiable at c and f(x) is differentiable at c, then does g(x) necessarily have to be?

(f*g)' = f'g + fg'

This shows that f(x) times g'(x) exists at c. I think it can be proven that this is not true, but I'll just provide a counter example:

Let f(x) = x^3 and g(x) = 1/x, and c = 0

(f(x)*g(x)) = x^3*1/x = x^2

(f(x)*g(x))' = 3x^2*1/x + x^3*-1/x^2 = 3x - x = 2x

(f(x)*g(x))'(0) = 0

g'(x) = -1/x^2

g'(0) = undefined!