I come across this theorem in a Advanced Calculus text. It concerns the chain rule of n variables. The theorem states:
Let $L$$\in $Lin$(\mathbb R^{p}, \mathbb R^{q}.$ Then $\ D \circ L$=$L \circ D,\ $that is, for every mapping $f:U \to \mathbb R^{p}\ $differentiable at $a\in$$U$ we have, if $U$ is open in $\mathbb R^{n},$
$$D(Lf)(a)=L(Df)(a)$$ Proof: Using the chain rule and the linearity of $L$ we find
$$D(L \circ f)(a)=DL(f(a)) \circ Df(a)=L \circ (Df)(a).$$
What I would like to know is, how did the author got the right hand side equality: in the proof $DL(f(a)) \circ Df(a)=L \circ (Df)(a).$
Thank you all in advance.
The differential of a linear map at a point is itself, so $DL(f(a))=L$.
Note that $L(c+h)-L(c)-L(h)=0$, and hence $\dfrac{|L(c+h)-L(c)-L(h)|}{|h|}=0\rightarrow 0$, so by uniqueness of differential, $DL(c)=L$.