Let $k$ be an algebraically closed field, and let $\mathbb{A}^n(k)$ denote the linear $n$-dimensional vector space over $k$. Suppose $X \subset \mathbb{A}^n$, $X' \subset \mathbb{A}^m$ are two varieties. We say that $X, X'$ are isomorphic if there exists a polynomial map $\phi : X \rightarrow X'$ which is bijective and whose inverse is a polynomial map. Note that $\phi$ is polynomial if $\phi(x) = (f_1(x), \cdots, f_m(x))$ where each $f_j$ is a polynomial. I want to show that if $p \in X$ is smooth (the tangent space of $X$ at $p$ has the same dimension as $X$) if and only if $\phi(p)$ is a smooth point of $X'$. It is not clear to me why if $(x_1, \cdots, x_n)$ is an eigenvector of the Jacobian of the generators of $I(X)$ at $p$, that $\phi((x_1, \cdots, x_n))$ must be an eigenvector of the Jacobian of the generators of $I(X')$, which I presume to be the natural approach to show that the two tangent spaces have equal dimension.
Any suggestions?