I am trying to prove the function is closed under multiplication (in a step to prove it's a subspace).
The problem is defined as:
Show that the set of differentiable real-valued functions $f$ on the interval $(-4,4)$ such that $f'(-1)=3f(2)$ is a subspace of $\mathbb{R}^{(-4,4)}$
Now to prove it is closed under multiplication, we show, that multiplying by $k \in \mathbb{R}$:
\begin{align} kf'(-1) &= k \, 3f(2) \\ (kf')(-1) &= 3 \, (kf)(2) \end{align}
How should I think about the second line above in terms of scaling the function? Basically what is it expressing? An explanation would be nice. Thanks.
Short answer $$\frac{d}{dx}(cf(x)) = c\frac{d}{dx}f(x)$$
I would do it likes this:
Let $f\in \{\text{your space}\}$ and $k\in \mathbb{R}$. To show that $kf\in \{\text{your space}\}$ we must show that $$(kf)'(-1)=3(kf)(2) \tag{1}$$ (Notice here that the derivative sign is outside $kf$).
As is well known (and easy to show) $$\frac{d}{dx}(cf(x))=c\frac{d}{dx}f(x) \tag{$c\in \mathbb{R}$}$$ so $(kf)'(-1)=k(f'(-1))$ which means that $(1)$ becomes $$k(f'(-1))=k(3f(2))$$ which holds for all non-zero $k$ since $$f'(-1)=3f(2)$$ for all $f\in \{\text{your space}\}$ and $k=0$ is trivial.