In calculus, the antiderivative (indefinite integral) can be considered as the reverse operation of a derivative.

A gradient yields a vector. Is there a similar way of reversing gradient, as you do with derivatives?

  • $\begingroup$ Haven't you seen line integrals, they yield vectors. $\endgroup$
    – L.K.
    Jun 15, 2014 at 7:16
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    $\begingroup$ @syockit "Reversing" a gradient shouldn't yield a vector, it should yield a scalar field. The gradient itself is a vector, but the function on which the gradient is applied is a scalar field. $\endgroup$
    – M. Vinay
    Jun 15, 2014 at 7:19
  • $\begingroup$ @M.Vinay Sorry for the confusing statement. I fixed the question. $\endgroup$
    – syockit
    Jun 15, 2014 at 12:46

4 Answers 4


Yes, it is possible. I will give an example to demonstrate the general procedure. Consider $f(x, y) = x^3 - 3xy^2 + x^2 + y^2 + \log x$. Then $\nabla f = \dfrac{\partial f}{\partial x}\hat{i} + \dfrac{\partial f}{\partial y}\hat{j} = \left(3x^2 - 3y^2 + 2x + \dfrac{1}{x}\right)\hat{i} + (2y -6xy)\hat{j}$.

To reverse this, we look at each component individually. We know that
$\dfrac{\partial f}{\partial x} = 3x^2 - 3y^2 + 2x + \dfrac{1}{x}\\ \dfrac{\partial f}{\partial y} = 2y - 6xy $


$\begin{align} \displaystyle f(x,y) & = \int \dfrac{\partial f}{\partial x}\, dx\\ & = \int 3x^2 - 3y^2 + 2x + \dfrac{1}{x}\, dx\\ & = x^3 - 3xy^2 + x^2 + \log x + u(y) \end{align}$

What is that $u(y)$? It's the "constant" of integration, of course.When we differentiate $f$ with respect to $x$ partially, any term of $f$ not containing $x$ is a constant - this includes terms containing only $y$.

Now to determine $u(y)$, we look at $\dfrac{\partial f}{\partial y}$. We could integrate this with respect to $y$, similar to what we did with $\dfrac{\partial f}{\partial x}$. Then some of the terms after the integration will be common. The terms that are not common are those that constitute $u(y)$. So we need to look for terms of $\dfrac{\partial f}{\partial y}$ that do not contain $x$, and integrate them. Here, $\dfrac{\partial f}{\partial y} = 2y - 6xy$, and the only term not containing $x$ is $2y$. Therefore:

$\displaystyle u(y) = \int 2y\, dy = y^2 + C$.

Thus, $\boxed{f(x, y) = x^3 - 3xy^2 + x^2 + \log x + y^2 + C}$.

In general: $$f(x, y) = \int \dfrac{\partial f}{\partial x} dx + \int \left[\text{terms of $\dfrac{\partial f}{\partial y}$ that do not contain $x$}\right]\, dy$$

  • $\begingroup$ On the third line of the derivation after "Therefore:", there is an erroneous integral sign. Otherwise, nice answer! Is there a general name for this procedure? $\endgroup$ Jun 15, 2014 at 8:39
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    $\begingroup$ @JohnvonN. Thanks! Hm, I don't know, but this is the same as the method for solving exact equations - equations of the form $\dfrac{\partial f}{\partial x}dx + \dfrac{\partial f}{\partial y}dy = 0$ (the solution of which is $f(x, y) = C$, where $f(x, y)$ is found using the above method). $\endgroup$
    – M. Vinay
    Jun 15, 2014 at 8:43
  • $\begingroup$ I came across something similar in Kreyszig's Advanced Engineering Mathematics, in a topic covering integrating factors for exact differential equations. $\endgroup$
    – syockit
    Jun 15, 2014 at 12:58
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    $\begingroup$ @kaka Of course. It's possible to recover a [differentiable] function from its derivative (except for the constant term), and it's of course possible to find the second derivative [if it's twice differentiable] without even finding the original function. Similarly, it's possible to recover a scalar field from its gradient, thanks to the method in the answer [i.e., by solving the "exact equation"]. $\endgroup$
    – M. Vinay
    Nov 26, 2016 at 9:11
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    $\begingroup$ @kaka Knowing the derivative at all points in an interval is [almost] as good as knowing the function at all points in the interval, since "differentiation" is invertible (except for the constant term). The same happens to be true of gradients of scalar fields too. $\endgroup$
    – M. Vinay
    Nov 26, 2016 at 9:13

I prefer a simpler one-step solution: Choose the curve $\vec\gamma(t) = t\vec x$ on $t\in[0,1]$, with endpoints $\vec\gamma(0)=\vec 0$ and $\vec\gamma(1) = \vec x$, and then the gradient theorem implies that $$\begin{align} f(\vec x) &= f(\vec 0) + \int_0^1 \nabla f(\vec\gamma(t))\cdot\vec\gamma'(t)\,\mathrm dt \\ &= f(\vec 0) + \int_0^1 \nabla f(t\vec x)\cdot\vec x\,\mathrm dt. \end{align}$$ The unknown $f(\vec 0)$ plays the role of the constant of integration.

For example, if we have $\nabla f(x,y)=\begin{bmatrix}3x^2 - 3y^2 + 2x\\2y - 6xy\end{bmatrix}$, then $$\nabla f(tx,ty)\cdot(x,y) = 3 t^2 (x^3-3xy^2)+2 t (x^2+y^2)$$ from which we easily get $$f(x,y)=f(0,0)+x^3-3xy+x^2+y^2.$$ The $\log x$ term in M. Vinay's answer complicates things a little as the function is undefined at $\vec 0$. It is straightforward to get around this by taking a different starting point, say $\vec x_0=(1,0)$ and choosing the curve $\vec\gamma(t)=\vec x_0+t(\vec x-\vec x_0)$.

  • $\begingroup$ This was a great answer, and IMHO more helpful than the others. I was not clear on the issue where $\nabla f(tx,ty)\cdot (x,y)$. Where does the $(x,y)$ come from? Is $\vec{x}$ presumed to be a two dimensional vector of $(x,y)$? That would make sense. $\endgroup$
    – ABC
    Dec 26, 2022 at 2:55

M. Vinay's answer is good. For more than 2 variables, the process can be chained. For $$F(x,y,z) = f_x x + f_y y + f_z z = \nabla f$$ we can integrate with respect to $x$

$$f(x,y,z)=\int f_x dx + g(y, z)$$

where $g$ is a function only in terms of $y$ and $z$. Then differentiating with respect to $y$

$$f_y(x,y,z)=\frac{\partial}{\partial y} \int f_x dx \ + g_y(y,z) $$

The first term is common and we are left with $g_y(y,z)$, which can be integrated to get $$g(y,z) = \int g_y(y,z) dy + h(z)$$

$$f(x,y,z)=\int f_x dx + \int g_y(y,z) dy + h(z)$$

Then repeating the process by differentiating $f$ with respect to $z$ and then integrating, we get all the terms of $f$:

$$f(x,y,z)=\int f_x (x,y,z) dx + \int g_y(y,z) dy + \int h_z(z) dz$$


In general, if the $\nabla$ operator is expressed in some orthogonal coordinates $\mathbf{q}=(q^1,q^2,q^3)$, the gradient of a scalar function $\varphi(\mathbf{q})$ will be given by $$\nabla \varphi(\mathbf{q})=\frac{\hat{\mathbf{e}}_{i}}{h_{i}} \frac{\partial \varphi}{\partial q^{i}} \tag{1}$$ And a line element will be $$d \boldsymbol{\ell}=h_{i} d q^{i} \hat{\mathbf{e}}_{i}\tag{2}$$ So the dot product between these two vectors is $$\nabla \varphi(\mathbf{q})·d \boldsymbol{\ell}=(\frac{\hat{\mathbf{e}}_{i}}{h_{i}} \frac{\partial \varphi}{\partial q^{i}} )·(h_{i} d q^{i} \hat{\mathbf{e}}_{i})=\frac{\partial \varphi}{\partial q^{i}}dq^i \tag{3}$$

Therefore, $\nabla \varphi(\mathbf{q})·d \boldsymbol{\ell}$ yelds the total derivative of $\varphi(\mathbf{q})$: $$\nabla \varphi(\mathbf{q})·d \boldsymbol{\ell}=d\varphi(\mathbf{q})\tag{4}$$

So you can recover the $\varphi(\mathbf{q})$ function at the right of the nabla operator just by evaluating the line integral of $(4)$: $$ \varphi(\mathbf{q}_2)-\varphi(\mathbf{q}_1) =\int_\gamma \nabla \varphi(\mathbf{q})·d \boldsymbol{\ell} \tag{5} $$


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