Vtyjkyuk

Corollary 3.46.
Suppose $G subset operatornameGL(n, mathbbC)$ is a matrix Lie group with Lie algebra $mathfrakg$.
Then a matrix $X$ is in $mathfrakg$ if and only if there exists a smooth curve $gamma$ in $operatornameM_n(mathbbC)$ with $gamma(t) in G$ for all $t$ and such that $gamma(0) = I$ and $mathrmdgamma/mathrmdt|_t=0 = X$.
Thus, $mathfrakg$ is the tangent space at the identity to $G$.

Proof.
If $X$ is in $mathfrakg$, then we may take $gamma(t) = e^tX$ and then $gamma(0) = I$ and $mathrmdgamma/mathrmdt|_t=0 = X$.
In the other direction, suppose that $gamma(t)$ is a smooth curve in $G$ with $gamma(0) = I$.
For all sufficiently small $t$, we can write $gamma(t) = e^delta(t)$, where $delta$ is a smooth curve in $mathfrakg$.
Now, the derivative of $delta(t)$ at $t = 0$ is the same as the derivative of $t mapsto t delta'(0)$ at $t = 0$.
Thus, by the chain rule, we have
$$
gamma'(0)
= left.
fracmathrmdmathrmdt e^delta(t)
right|_t=0
= left.
fracmathrmdmathrmdt e^t delta'(0)
right|_t=0
= delta'(0).
$$
Since $delta(t)$ belongs to $mathfrakg$ for all sufficiently small $t$, we conclude (as in the proof of Theorem 3.20) that $delta'(0) = gamma'(0)$ belongs to $mathfrakg$.

(Original image here.)

The conclusion say $delta in mathfrakg implies delta'(0) in mathfrakg$. So why does the derivative belong to $frakg$ too?

Since Lie algebra is a linear space, we suppose $v_1,v_2,cdots,v_n$ is a basis of $mathfrakg$.

$delta(t)in mathfrakg$, so we have $delta(t)=sum_j=1^nf_j(t)v_j$.

It's obvious that $delta'(t)|_t=0=sum_j=1^nf_j'(0)v_jin mathfrakg$.