Why do derivatives pop up in abstract algebra? Why is it that derivatives, which are used in calculus, are also useful in abstract algebra concerning abstract polynomial rings? What is the connection between calculus and algebra here? 
 A: Partial derivatives can give information about "functional dependence".
Here is a fact that is easy for polynomials over $\mathbb{R}$ but is true in general for polynomials over any  perfect field.
Let $\phi_1$, $\ldots$, $\phi_n$ be $n$ polynomials in $n$ variables $x_1$, $\ldots$, $x_n$ over a field $F$. Assume that the jacobian determinant $\det (\frac{\partial \phi_i}{\partial x_j})$ is not the zero polynomial. Then the $\phi_i$'s are functionally independent, that is, there does not exist a non-zero polynomial $P$ such that $P(\phi_1, \ldots, \phi_n ) \equiv 0$.
Assume such $P$ there exists and take one of minimal degree. From the equality  $P(\phi_1, \ldots, \phi_n ) = 0$ we get a system of equations in $\frac{\partial P}{\partial y_i}(\phi_1, \ldots, \phi_n)$:
$$\sum_i \frac{\partial P}{\partial y_i}(\phi_1, \ldots, \phi_n)\cdot \frac{\partial \phi_i}{\partial x_j}= 0$$
$j=1,\ldots, n$ and so by Cramer
$$\det (\frac{\partial \phi_i}{\partial x_j}) \cdot \frac{\partial P}{\partial y_k}(\phi_1, \ldots, \phi_n)=0$$
for all $k = 1,\ldots, n$ and since the jacobian is nonzero and the ring of polynomials is a domain, we get $\frac{\partial P}{\partial y_k}(\phi_1, \ldots, \phi_n)\equiv 0$ for all $k=1,\ldots, n$. Since $P$ was chosen of minimal degree we get $\frac{\partial P}{\partial y_k}(y_1, \ldots, y_n) \equiv 0$, and so $P$ is either a constant polynomial (in the case char $F = 0$) or a polynomial in $y_i^p$ if char $F= p$. Now since $F$ is perfect, in the second case we necessarily have $P = Q^p$ for some $Q \in F[y_1, \ldots, y_n]$. In either case we get a contradiction.
Note that if we were working over $\mathbb{R}$ this meant that around any point where the jacobian is not $0$ the map $(x_1, \ldots, x_n) \mapsto (\phi_1(x), \ldots, \phi_n(x))$ is a local diffeomorphism (analytic in fact) so the image of the map $\phi$ contains an open set, hence it is "Zariski dense".
$\bf{Added:}$ Let $R_1$, $\ldots$, $R_{k+1}$ rational functions of several variables such that the jacobian matrices of $(R_1, \ldots, R_k)$ and $(R_1, \ldots, R_{k+1})$ have the same rank. Then $R_{k+1}$ is algebric over $(R_1,\ldots, R_k)$. Indeed, otherwise there would exist an $F$ derivation of the field $F(x_i)$ that is $0$ on $R_1$, $\ldots$, $R_k$, but $1$ on $R_{k+1}$, but this is not possible, because of the rank condition on the jacobians.
