I want to prove differential of det map at some matrix $A$ is given by $f(A)\operatorname{tr}(X)$
let $f:GL_n(\mathbb{R})\rightarrow\mathbb{R}$ is det map i.e $f(A)=\det A$,
claim: $f_{*}(AX)=(f(A))\operatorname{tr}(X)$
proof of claim:
Let a curve $c(t)=Ae^{tX}$ such that $c(0)=A,c'(0)=AX$. using this curve we calculate differential
$f_{A{*}}(AX)=\frac{d}{dt}|_{t=0}f(c(t))=c'(t)|_{t=0}f(c(t))|_{t=0}=AXf(A)=AX(\det A)$, where am I getting wrong? Thank you.