Would love some feedback on my attempt at this solution: If $f: A⊂ R^n → R^m$ and $g: B ⊂ R^n → R^m$ are differentiable functions on the (open) sets A and B and α, β are constants, prove that $αf + βg: A ∩ B ⊂ R^n → R^m$ is differentiable and $D(αf + βg)(x) = αDf(x) + βDg(x)$.
f and g are differentiable on A,B. Thus, $∀ε > 0, ∃ δ1 > 0 s.t. ||x – x_0|| < δ_1 ⇒ α||f(x) – f(x_0) – Df(x_0)(x – x_0)|| ≤ ε/α ||x – x_0|| ≤ ε_0/2 ||x – x_0|$
and $β||g(x) – g(x_0) –Dg(x_0)(x – x_0)|| ≤ε/β||x – x_0|| ≤ε_0/2 ||x – x_0||$ when $||x – x_0|| < δ_2$ where $ε/α , ε/β ≤ ε_0/2$ .
Then we check if αf + βg is differentiable. Take $δ_0 = min{δ_1,δ_2}$. Then $||(αf + βg)(x) – (αf + βg)(x_0) – (αDf(x_0) + βDg(x_0))(x – x_0)|| = ||(αf(x) + βg(x)) - αf(x_0) - βg(x_0) – (αDf(x_0) + βDg(x_0))(x-x_0)|| ≤ α||f(x) – f(x_0) –Df(x_0)(x – x_0)|| + β||g(x) – g(x_0) – Dg(x_0)(x – x_0)|| ≤ ε_0/2 ||x – x0|| + ε_0/2 ||x – x_0|| = ε_0||x – x_0||$ when $||x – x_0|| ≤ δ_0$.
There is some hassle with $\epsilon/\alpha$, etc. We also have to take care of the cases $\alpha=0$ or $\beta=0$.
Write: Given an $\epsilon>0$ there is a $\delta_1>0$ such that $\|x-x_0\|<\delta_1$ implies $$\|f(x)-f(x_0)-Df(x_0).(x-x_0)\|\leq{\epsilon\over 2(|\alpha|+1)}\|x-x_0\|\ ,$$ and similarly for $g$. Put $h:=\alpha f+\beta g$. Then $$\bigl\|h(x)-h(x_0)-\bigl(\alpha Df(x_0)+\beta Dg(x_0)\bigr).(x-x_0)\bigr\|\leq\epsilon\|x-x_0\|\ ,$$ as soon as $\|x-x_0\|<\delta:=\min\{\delta_1,\delta_2\}$.