The usual thing one does (but this can vary according to what you're trying to achieve) is to compute a vector called the gradient of the function, notated $\nabla f$. The gradient has one component for each input to $f$; for each component you treat all other variables temporarily as constants and differentiate with respect to the chosen one:
$$(\nabla f)(x,y) = \left(\frac{d}{dx}f(x,y), \frac{d}{dy}f(x,y)\right)
= ( 2(x+y), 2(x+y) ) $$
In this example it may be a bit hard to see what's going on, because both components end up being the same, so let's try another one: $g(x,y)=x^2y+y$ gives
$$\nabla g(x,y) = (2xy, x^2+1)$$
The intuitive property of the gradient is that if we have some small offset $(h,k)$ from $(x,y)$, then the difference $f(x+h,y+k)-f(x,y)$ is closely approximated by the dot product between $(h,k)$ and $\nabla f(x,y)$.
The components of the gradient are called partial derivatives and are often notated with a special sign as $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. The $\partial$ sign looks a bit like a $d$, but is different as a reminder that a single partial derivative is not usually enough to estimate $\Delta f$; you need both of them.
No, differentiating once gives you the differential, a row vector $Df(x)$.
The transpose of that vector is the gradient, $\nabla f(x)=Df(x)^\top$ (except in physics, numerics, optimization, where the derivative is sometimes denoted with the inverted triangle). Note that in generalized situations, transposition involves the coefficients of the scalar product.
Differentiating twice gives the second derivative, which is a tensor of order 2. Its coefficents can be arranged in matrix form, this is then called the Hesse or Hessian matrix.
The higher derivatives have no extra name.
First and second derivative are frequently used to form a quadratic approximation of the function. This is useful to compute directions of descend and testing for local minimality.
Best Answer
In $n$ dimensions, $\nabla$ is the operator $$\nabla = \left(\frac{\partial}{\partial x_1},\cdots,\frac{\partial}{\partial x_n}\right)$$that acts on a differentiable function $f: \Bbb R^n \to \Bbb R$ producing: $$\nabla f = \left(\frac{\partial f}{\partial x_1},\cdots,\frac{\partial f}{\partial x_n}\right).$$Make $n= 1$. So: $$\nabla = \left( \frac{\partial}{\partial x}\right) = \left( \frac{\rm d}{{\rm d}x}\right) = \frac{\rm d}{{\rm d}x},$$ since we identify $1 \times 1$ matrices with numbers. If $f(x) = x^2$, we have: $$\nabla f = 2x = f'(x).$$