Prove that f: $\Bbb R^2$-> $\Bbb R$, (x,y)$\mapsto$ x$^2$+ 2xy$^2$ +5y$^3$ is differentiable at (2,1) with DF(2,1)=[6,3].
Now I know that the partial derivatives
1) $\partial f/\partial x (2,1)=2x + 2y^2$ and when x=2,y=1 I get 6
2)$\partial f/\partial y (2,1)=4xy + 15y^2$ and when x=2,y=1 I get 23
However, after this, I'm not very sure where to go with that... The book that i'm going off of says this for DF
"Let V = V1 × ... × Vd (V1,...,Vd being Banach spaces), Ω open in V, a = $(a^1,...,a^d)$ ∈ Ω. A function f from Ω to some Banach space is said to be partially differentiable at a in the j-th variable (j ∈ {1, . . . , d}) if the function $x^j$ → $f(a^1,...,a^j−1,a^j + x^j,a(j+1),...,a^d)$ is differentiable at $x^j$ = 0. The corresponding derivative is then denoted by D$_j$f(a) and called the j-th partial derivative of f at a."
Thanks for the help _BradB
In the case $V_i = \Bbb R$ for all $i$, and the codomain being $\Bbb R^n$, for some $n$, i.e., $f: \Bbb R^d \to \Bbb R^n$, $x = (x_1,...,x_d) \mapsto f(x) = (f_1(x),...,f_n(x))$, if $a \in \Bbb R^d$ is a point at which $f$ is differentiable, then $Df(a)$ is simply the Jacobian matrix of $f$ at $a$. That's,
$$Df(a) = \left( \frac{\partial f_i}{\partial x_j}(a) \right)_{i,j} = \begin{pmatrix} \frac{\partial f_1}{\partial x_1}(a) & \ldots & \frac{\partial f_1}{\partial x_d}(a) \\ \vdots & \ddots & \vdots \\ \frac{\partial f_n}{\partial x_1} (a) & \ldots & \frac{\partial f_n}{\partial x_d} (a) \end{pmatrix}$$
In the case where $n = 1$, i.e. $f: \Bbb R^d \to \Bbb R$, $Df(a)$ simplifies to a row matrix which is the transpose of the gradient of $f$ at $a$; i.e. $Df(a) = \nabla f(a) ^T$, with:
$$\nabla f(a) = \begin {pmatrix} \frac{\partial f}{\partial x_1}(a) \\ \vdots \\ \frac{\partial f}{\partial x_d}(a) \end{pmatrix}$$
In your example, $f$ has partial derivatives everywhere and they are continuous at $(2,1)$. Hence, $f$ is differentiable, and,
$$Df(2,1) = (\frac{\partial f}{\partial x}(2,1), \frac{\partial f}{\partial y}(2,1)) = (6,23)$$