I asked the following question on math.stackexchange but no one seemed to have an authorative answer so I'm posting here hoping that experts will see it.

The gradient is a tensor $\nabla f:\mathbf{V} \to \mathbf{R}$ where the partial derivatives are evaluated at some point $(x_0, y_0, z_0)$. And evaluation of this linear form at some vector $v=(v_1,v_2,v_3)$ gives

$$ (\nabla f)(\mathbf{v}) = \partial_x f v_1 + \partial_y f v_2 + \partial_z f v_3 $$ Furthermore in going to a new coordinate system these partial derivatives transform in the expected way.

But what about a function $g:\mathbf{V} \to \mathbf{R}$ which is defined only using the partial derivative of $f$ in the $x$ direction. $$ g(\mathbf{v}) = \partial_x f v_1 + \partial_x f v_2 + \partial_x f v_3 $$ As I understand this is not considered a tensor because in moving to a new coordinate system it does not transform correctly.

This has confused me endlessly. The strict definition considers a map such as $f:\mathbf{V}\times\mathbf{V} \to \mathbf{R}$ a tensor if linearity holds in each parameter. The function $g$ above certainly satisfies that. It seems to me that this definition is not used and that the definition of a tensor that is actually used consists of two parts.

• linearity in each parameter (i.e. multilinear form),
• and the algebraic structure of the coefficients is maintained in coordinate transformation

Because once we have calculated $\partial_x f$ it is just a scalar and we just hit $(\partial_x f,\partial_x f,\partial_x f)$ with the usual transformation for a covariant vector to get the new coefficients for $g$ in the new coordinate system. That these new coefficients don't have the right algebraic structure doesn't make multilinearity of $g$ go away. Is this at all correct?