Proving Total Derivative of jf+kg at a

[JohnLeee]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

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)?

 

[HallsofIvy]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

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

What does this mean? Since f and g are from Rⁿ to R^m, they are functions of n variables. In order to talk about "x+ h", you have to be thinking of x and h as vectors in Rⁿ. But then division by h is division by a vector and that is not defined. You probably mean to divide by |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)?

But if you do divide by |h| you still to have to deal with the fact that h can go to 0 along many different paths.

 

[JohnLeee]

So should I approach this problem completely differently?