## Totally Differentiable?

Q. If U⊆R^n is an open set with a ∈ U, and f: U->R^m and g: U->R^m are totally differentiable at a, prove that jf+kg is also totally differentiable at a and that (D(jf+kg))a = j(Df)a+k(Dg)a.

What I did: Let p(x) = jf(x)+kg(x)

Then [p(x+h)-p(x)]/h = [jf(x+h)+kg(x+h)-jf(x)-kg(x)]/h
= j[f(x+h)-f(x)]/h + k[g(x+h)-g(x)]/h

Then I took the limit of that as h goes to 0. Since both terms exist and are 0, jf+kg is also differentiable.

Is this the right way to solve the problem? I'm not sure because of the term "totally" differentiable. Also how do I solve the second part of the problem (total derivative)?

ps: does this topic go to the Analysis forum or stay here?