Lie operator is left exact In page 190, part (c) of the book "Algebraic groups, the theory of group schemes of finite type over a field" of Milne, there's a part stating that:

An exact sequence of algebraic groups $e \rightarrow G' \rightarrow G \rightarrow G''$ gives an exact sequence of Lie algebras $0 \rightarrow Lie(G') \rightarrow Lie(G) \rightarrow Lie(G'')$.

This is what I tried so far: Because every algebraic group is affine over some base field, the exact algebraic groups sequence gives the exact Hopf algebras sequence $0 \rightarrow A' \rightarrow A \rightarrow A''$. If we can obtain from this sequence the sequence $0 \rightarrow I_A/I_A^2 \rightarrow I_{A'}/I_{A'}^2 \rightarrow I_{A''}/I_{A''}^2$, then we are done, because $Lie(G) \cong Hom(I_G/I_G^2, k)$, and the Hom functor is left exact.
But I can't prove that missing point. I would be glad if anyone can give me any hint. Thanks a lot.
Edited: From the hint of @anon, I post here where I stuck right now:
Because we have an exact sequence $$0 \rightarrow Lie(G) \rightarrow G(k[\epsilon]) \rightarrow G(k)$$ and $G$ is an left-exact functor, we have this diagram:

Here all the rows are exact due to what we state above, and two last columns are exact due to left-exactness of G. So we need to prove that the first column is also exact. I think maybe this is some property from commutative algebra related to five lemma, but I couldn't prove it.

By definition, there is an exact sequence
$0\to\text{Lie}(G)\to G(k[\varepsilon]) \to G(k)$. For each $k$-algebra $R$, the functor $G \mapsto G(R)$ is left exact, from which the statement follows by a diagram chase in the obvious diagram.
