Let $G\subseteq\text{Mat}_{n\times n}(\mathbb{C})$ be a linear algebraic group and be given as the zero locus of an ideal $J\subseteq\mathbb{C}[x_{ij}\mid i,j\in [n]]$. In this case, the Lie algebra of $G$ is defined as $$ Lie(G)=\{X\in \text{Mat}_{n\times n}(\mathbb{C})\mid e^{tX}\in G\text{ for all }t\in\mathbb{R}\}.$$
It is known that $Lie(G)$ is equal to the tangent space of $G$ at the identity element, i.e. $$ Lie(G) = T_I(G) = \{X\in\text{Mat}_{n\times n}(\mathbb{C})\mid df\mid_I(X)=0\text{ for all }f\in J\}. $$
I know that there are standard proofs, but I wanted to try to prove it myself. One direction was straightforward but I am not sure about the other direction. I came up with the following and I would like to know if the proof is correct:
$Lie(G)\subseteq T_I(G):$ Let $X\in Lie(G)$ and let $f\in J$. I note that $$ f(I-\varepsilon X)=f(I)-\varepsilon df\mid_I(X) +\text{terms with }\varepsilon^2. \quad (*) $$
On the other hand, for $\varepsilon$ is small enough (say for $|\varepsilon|<\delta$), I have $\Vert I-\varepsilon X\Vert_F<1$. Then, $\log(I-\varepsilon X)$ is well defined and I can write $$ f(I-\varepsilon X)=f(e^{\log(I-\varepsilon X)})=f(e^{\sum_{m=1}^\infty\frac{\varepsilon^m X^m}{m}})=f\Big(\sum_{m=0}^\infty\frac{(\varepsilon X+\varepsilon^2 X^2/2 +...)^m}{m!}\Big) =f(I+\varepsilon X+\varepsilon^2 Y) $$ for some matrix $Y$. Now, I have $f(I-\varepsilon X)=f(I+\varepsilon X+\varepsilon^2 Y)=f(I)+\varepsilon df\mid_I(X)+$terms with $\varepsilon^2$. Using $(*)$ and the fact that the equalities hold for all $|\varepsilon|<\delta$, I deduce that $df\mid_I(X)=-df\mid_I(X)$ so $df\mid_I(X)=0$. Thus, $X\in T_I(G)$.