[proofplan]
We compare three local quantities at $p$: the rank of the [Jacobian matrix](/page/Jacobian%20Matrix), the dimension of the Zariski tangent space, and the Krull dimension of the local ring. The generators $f_1,\dots,f_r$ of the full vanishing ideal cut out the tangent space by their first-order parts, so $\dim_k T_pX=n-\operatorname{rank}J(f_1,\dots,f_r)_p$. The hypothesis on irreducible components says that the local dimension of $X$ at $p$ is $d$. Finally, over an [algebraically closed field](/page/Algebraically%20Closed%20Field), smoothness at $p$ is equivalent to regularity of the local ring, i.e. equality between embedding dimension and local dimension.
[/proofplan]
[step:Translate the point into the local coordinate ring]
Let
\begin{align*}
S:=k[x_1,\dots,x_n],
\end{align*}
and let
\begin{align*}
A:=S/I(X)
\end{align*}
be the coordinate ring of $X$. Since $p=(a_1,\dots,a_n)\in X$, define the maximal ideal of $S$ corresponding to $p$ by
\begin{align*}
\mathfrak n_p:=(x_1-a_1,\dots,x_n-a_n)\trianglelefteq S.
\end{align*}
Because every element of $I(X)$ vanishes at $p$, we have $I(X)\subset \mathfrak n_p$. Hence
\begin{align*}
\mathfrak m_p:=\mathfrak n_p/I(X)\trianglelefteq A
\end{align*}
is the maximal ideal corresponding to $p$. Let
\begin{align*}
R:=A_{\mathfrak m_p}
\end{align*}
be the local ring of $X$ at $p$, and let
\begin{align*}
\mathfrak m_R:=\mathfrak m_pR
\end{align*}
be its maximal ideal.
The quotient map $A\to R$ induces a natural $k$-linear isomorphism
\begin{align*}
\mathfrak m_p/\mathfrak m_p^2 \cong \mathfrak m_R/\mathfrak m_R^2.
\end{align*}
Indeed, localization at $\mathfrak m_p$ does not change the residue field, which remains $A/\mathfrak m_p\cong k$, and the usual localization isomorphism gives
\begin{align*}
\mathfrak m_R/\mathfrak m_R^2\cong (\mathfrak m_p/\mathfrak m_p^2)\otimes_A A_{\mathfrak m_p}.
\end{align*}
Since $\mathfrak m_p$ acts by zero on $\mathfrak m_p/\mathfrak m_p^2$, this [tensor product](/page/Tensor%20Product) is naturally the same $k$-[vector space](/page/Vector%20Space) as $\mathfrak m_p/\mathfrak m_p^2$.
[/step]
[step:Compute the tangent space from the Jacobian matrix]
Define the Zariski tangent space at $p$ by
\begin{align*}
T_pX:=\operatorname{Hom}_k(\mathfrak m_p/\mathfrak m_p^2,k).
\end{align*}
We claim that
\begin{align*}
\dim_k T_pX=n-\operatorname{rank}J(f_1,\dots,f_r)_p.
\end{align*}
Let
\begin{align*}
\theta:k^n\to T_p\mathbb A_k^n
\end{align*}
be the standard identification sending $v=(v_1,\dots,v_n)$ to the $k$-[linear map](/page/Linear%20Map)
\begin{align*}
\theta(v):\mathfrak n_p/\mathfrak n_p^2\to k
\end{align*}
defined by
\begin{align*}
\theta(v)(x_j-a_j+\mathfrak n_p^2)=v_j
\end{align*}
for each $1\le j\le n$. Under this identification, a vector $v=(v_1,\dots,v_n)\in k^n$ lies in $T_pX$ precisely when every $f\in I(X)$ has zero first-order part on $v$:
\begin{align*}
\sum_{j=1}^n \frac{\partial f}{\partial x_j}(p)v_j=0.
\end{align*}
Since $I(X)=(f_1,\dots,f_r)$, it is enough to impose these equations for $f_1,\dots,f_r$. Thus $T_pX$ identifies with the kernel of the $k$-linear map
\begin{align*}
L:k^n&\to k^r
\end{align*}
whose matrix in the standard bases is $J(f_1,\dots,f_r)_p$. Therefore rank-nullity gives
\begin{align*}
\dim_k T_pX=n-\operatorname{rank}J(f_1,\dots,f_r)_p.
\end{align*}
[guided]
The point of this step is to make the Jacobian matrix appear from first-order algebra. The tangent space to [affine space](/page/Affine%20Space) at $p$ is represented by linear functionals on
\begin{align*}
\mathfrak n_p/\mathfrak n_p^2,
\end{align*}
where
\begin{align*}
\mathfrak n_p=(x_1-a_1,\dots,x_n-a_n)\trianglelefteq k[x_1,\dots,x_n].
\end{align*}
The classes of $x_1-a_1,\dots,x_n-a_n$ form a $k$-basis of $\mathfrak n_p/\mathfrak n_p^2$, so a tangent vector is the same thing as a tuple $v=(v_1,\dots,v_n)\in k^n$, where $v_j$ is the value of the tangent functional on $x_j-a_j$.
Now impose the condition that the tangent vector is tangent to $X$. Since $X$ is defined by the full vanishing ideal $I(X)$, a tangent vector to $X$ must annihilate the first-order part of every function in $I(X)$. For any $f\in I(X)$, Taylor expansion modulo $\mathfrak n_p^2$ gives
\begin{align*}
f-f(p)\equiv \sum_{j=1}^n \frac{\partial f}{\partial x_j}(p)(x_j-a_j)\pmod{\mathfrak n_p^2}.
\end{align*}
Because $p\in X$ and $f\in I(X)$, we have $f(p)=0$. Evaluating the corresponding tangent functional determined by $v$ gives the linear condition
\begin{align*}
\sum_{j=1}^n \frac{\partial f}{\partial x_j}(p)v_j=0.
\end{align*}
It remains to reduce from all $f\in I(X)$ to the chosen generators. Since $I(X)=(f_1,\dots,f_r)$, every $f\in I(X)$ can be written as
\begin{align*}
f=\sum_{i=1}^r h_if_i
\end{align*}
with $h_i\in k[x_1,\dots,x_n]$. Differentiating and evaluating at $p$ gives
\begin{align*}
\frac{\partial f}{\partial x_j}(p)=\sum_{i=1}^r h_i(p)\frac{\partial f_i}{\partial x_j}(p)+\sum_{i=1}^r f_i(p)\frac{\partial h_i}{\partial x_j}(p).
\end{align*}
The second sum is zero because $f_i(p)=0$ for every $i$. Hence the first-order condition for all elements of $I(X)$ is generated by the first-order conditions for $f_1,\dots,f_r$.
Therefore $T_pX$ is exactly the kernel of the linear map
\begin{align*}
L:k^n&\to k^r
\end{align*}
whose matrix is
\begin{align*}
J(f_1,\dots,f_r)_p=\left(\frac{\partial f_i}{\partial x_j}(p)\right)_{1\le i\le r,\ 1\le j\le n}.
\end{align*}
By rank-nullity,
\begin{align*}
\dim_k T_pX=\dim_k\ker L=n-\operatorname{rank}L=n-\operatorname{rank}J(f_1,\dots,f_r)_p.
\end{align*}
[/guided]
[/step]
[step:Identify the local dimension with the component dimension through $p$]
The irreducible components of $X$ containing $p$ correspond to the minimal prime ideals of $A$ contained in $\mathfrak m_p$. The Krull dimension of the local ring
\begin{align*}
R=A_{\mathfrak m_p}
\end{align*}
is the maximum of the dimensions of those irreducible components. By hypothesis, that maximum is $d$. Hence
\begin{align*}
\dim R=d.
\end{align*}
[/step]
[step:Convert the rank condition into regularity of the local ring]
The embedding dimension of the local ring $R$ is
\begin{align*}
\operatorname{edim}R:=\dim_k \mathfrak m_R/\mathfrak m_R^2.
\end{align*}
By the natural isomorphism
\begin{align*}
\mathfrak m_R/\mathfrak m_R^2\cong \mathfrak m_p/\mathfrak m_p^2
\end{align*}
and the definition of $T_pX$, we have
\begin{align*}
\operatorname{edim}R=\dim_kT_pX.
\end{align*}
From the tangent-space computation,
\begin{align*}
\operatorname{edim}R=n-\operatorname{rank}J(f_1,\dots,f_r)_p.
\end{align*}
Since $\dim R=d$, the condition
\begin{align*}
\operatorname{rank}J(f_1,\dots,f_r)_p=n-d
\end{align*}
is equivalent to
\begin{align*}
\operatorname{edim}R=d=\dim R.
\end{align*}
By the local algebra criterion for regular local rings, a Noetherian local ring is regular precisely when its embedding dimension equals its Krull dimension. Therefore the Jacobian rank condition is equivalent to $R=\mathcal O_{X,p}$ being a regular local ring.
[/step]
[step:Translate regularity back into smoothness at $p$]
Because $k$ is algebraically closed, it is perfect. For affine varieties over a perfect field, a point is smooth precisely when its local ring is regular. Applying this smoothness-regularity criterion to the local ring
\begin{align*}
\mathcal O_{X,p}=R
\end{align*}
shows that $p$ is smooth if and only if $R$ is regular. By the previous step, $R$ is regular if and only if
\begin{align*}
\operatorname{rank}J(f_1,\dots,f_r)_p=n-d.
\end{align*}
This proves the claimed equivalence.
[/step]
