What is the meaning of "gradient with respect to $\mathbf x$"?


I am talking about the symbol $$\nabla_\mathbf x$$

Does that simply mean derivative with respect to $\mathbf x$?

  • $\begingroup$ I can think of two things... the gradient of the function at a point $x$, or the covariant derivative evaluated on the direction of a given vector $x$. $\endgroup$ – Aloizio Macedo Oct 3 '15 at 3:16
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    $\begingroup$ You need to give some context. That subscript $x$ may have different meanings. $\endgroup$ – Giuseppe Negro Oct 5 '15 at 23:14
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    $\begingroup$ I can't see neither the $\nabla_x$ symbol nor the expression (these exact words 'gradient with respect') in the linked Wikipedia page. It's also absent in the version from March Where did you find it? $\endgroup$ – CiaPan Oct 6 '15 at 8:46

You need to be careful with any notation since often authors define and use symbols in different manner. Although in this particular case, my mathematical experience tells me that most common are:

1. "Gradient's variable":
following Ruben Tobar's answer, it may be notation providing information with respect to which variables the whole gradient should be taken. To be more precise: if $x_i\in \mathbb{R}$ for $i= 1,...,n$ and $\mathbf{x}\in\mathbb{R}^n$ is a vector defined as $\mathbf{x}=\{x_1,...,x_n\}$, given differentiable $f(\mathbf{x}):\mathbb{R}^n\rightarrow\mathbb{R}$ we have: $$\nabla_{\mathbf{x}}f=(\partial_{x_1}f,...,\partial_{x_n}f)$$ Therefore $\nabla_{\mathbf{x}}f$ is a vector whose entities $\partial_{x_i}f$ are just standard derivatives of multi-variable function $f$ with respect to the real variable $x_i$ (so called partial derivatives). Note that I did not write here explicitly dependence of $f$ on $\mathbf{x}$ since it would be redundant and erroneous, but nevertheless both left hand side and all the entities on the right hand side are functions dependent on $\mathbf{x}$.
To elaborate about the notation a little: it is quite commonly used in partial differential equations, where you have multiple different variables and not to confuse your reader, when you use (or define) a differential operator you stress the variables used in definition of this operator. For example consider the following: take $\mathbf{x}\in{\mathbb{R}^n}$ defined as above and similarly $\mathbf{z}\in{\mathbb{R}^d}$ for some natural $r,d>1$ and $t\in\mathbb{R}$. Problem is to find a function $f(t,\mathbf{x},\mathbf{z}):\mathbb{R}\times \mathbb{R}^n\times\mathbb{R}^d\rightarrow\mathbb{R}$ solving differential equation. Without stressing the variables the following differential equation is ambiguous and difficult to write in any different way: $$f_t + \nabla_\mathbf{x}f +{\rm div}_\mathbf{z}f = 0$$ where ${\rm div}_\mathbf{z}$ is different differential operator (called divergence) defined as $\sum_i\partial_{z_i}$, $i=1,...d.$

2. Directional derivative:
As you see in https://en.wikipedia.org/wiki/Directional_derivative (where you can find more information) it is a very common notation for directional derivative. Not in full generality: Given any $f(\mathbf{x}):\mathbb{R}^n\rightarrow\mathbb{R}$ and a vector $\mathbf{v}\in \mathbb{R}^n$: $$\nabla_{\mathbf{v}} f(\mathbf{x}):=\nabla f(\mathbf{x})\cdot\mathbf{v},$$ where on the right hand side we have standard scalar product.

3. Gradient at x?:
In my opinion (some may disagree): it is neither common, nor useful notation. Although, using our notation it would mean: $$\nabla_\mathbf{x}f:=[\nabla f](\mathbf{x}):=(\partial_{x_1} f(\mathbf{x}),...,\partial_{x_n} f(\mathbf{x}))$$ Where $\mathbf{x}\in\mathbb{R}^n$ is fixed and partial derivatives of $f$ are evaluated at $\mathbf{x}$. Note that unlike in point 1. where we didn't stress dependence on the variable, here we write explicitly that functions are evaluated at $\mathbf{x}.$

  • $\begingroup$ To be clear about your differential equation in point 1, the first and last terms are scalar, and the second term takes values in $\mathbb{R}^n$, correct? $\endgroup$ – kdbanman May 26 at 21:49

Let $f=f(\boldsymbol x,\boldsymbol u)=f(x_1,...,x_n,u_1,...,u_r)$. Then

$$\nabla f=\left(\frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n},\frac{\partial f}{\partial u_1},...,\frac{\partial f}{\partial u_r} \right),$$ $$\nabla _{\boldsymbol x}f=\left(\frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n} \right),$$

and $$\nabla _{\boldsymbol u}f=\left(\frac{\partial f}{\partial u_1},...,\frac{\partial f}{\partial u_r} \right).$$ so that $\nabla f(\boldsymbol x,\boldsymbol u) = (\nabla_{\mathbf{x}}f(\boldsymbol x,\boldsymbol u),\nabla_{\mathbf{u}}f(\boldsymbol x,\boldsymbol u))$


I have this in my classical mechanics' notebook:

If $\mathrm{x}=(x_1,x_2,\ldots,x_n)$ then $$\nabla_{\mathrm{x}}f=\left(\partial_{x_1}f,\partial_{x_2}f,\ldots,\partial_{x_n}f\right)$$ Where $x_i$ are variables.

Hope it help you :)


Gradient simply means 'slope', and you can think of the derivative as the 'slope formula of the tangent line'. So yes, gradient is a derivative with respect to some variable.

In vector analysis, the gradient of a scalar function will transform it to a vector.

By definition, if $\phi$ is a scalar function of $\phi(x,y,z)$, then $\nabla\phi=\frac{\partial \phi}{\partial x}\mathbf{i} + \frac{\partial \phi}{\partial y}\mathbf{j} + \frac{\partial \phi}{\partial z}\mathbf{k}$.

As an example, if $\phi(x,y,z) = x^2y^3z$, then $\nabla \phi = 2xy^3z \mathbf{i} + 3x^2y^2z \mathbf{j}+x^2y^3\mathbf{k}$.

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    $\begingroup$ but you didn't answer my question. what is ∇_x ? so in above, ∇_x ϕ would be, 2xy^3*z? $\endgroup$ – user128616 Mar 11 '15 at 3:36
  • $\begingroup$ I believe it would read as … 'gradient AT $x$'. That is my opinion. So the resulting function would still be a vector, but evaluated AT $x$. $\endgroup$ – cgo Mar 11 '15 at 5:17

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