The total derivative appears similar to the directional derivative. I have a question on the similarity. I provide some up front definitions and then questions.

Directional Derivative Wiki of $f(x) = f(x_1, ..., x_n)$ is:

$\triangledown_v f(x) = \triangledown f(x) \cdot v = \lim_{h \rightarrow 0} \frac {f(x + hv) - f(x)} h$

where RHS is the dot product of the gradient and directional vector $v = (v_1, ..., v_n)$

Total Derivative Wiki as a differential form where $f: \mathbb{R}^n \rightarrow \mathbb{R}$ and $\Delta x = (\Delta x_1, ..., \Delta x_n)^T$, the total derivative of $f$ at $a$ is $d_{f_a} = (\frac {\partial f} {\partial x_1} (a), ..., \frac {\partial f} {\partial x_n} (a)) = \triangledown f(a)^T$. Then Wiki later continues by stating $f(a + \Delta x) - f(x) \approx df_a \cdot \Delta x$, then goes on by redefining $d_{f_a} = \sum_{i=1}^n \frac {\partial f} {\partial x_i} (a) d x_i$.


  1. It appears from these definitions the total derivative is reduced to the gradient. Is this true?

  2. $d_{f_a}$ is later redefined as the dot product of $\triangledown f(x)^T \cdot \Delta x$. This appears to be the directional derivative. Is this correct or an error? Is the direction vector a scaling of a unit vector in every direction?


1 Answer 1

  1. It appears from these definitions the total derivative is reduced to the gradient. Is this true?

The gradient is a vector. The total derivative, on the other hand, is the linear map $df_a \colon \Bbb{R}^n \to \Bbb{R}$ defined by taking the dot product with the gradient: $$ df_a(v) = \nabla f(a) \cdot v = \nabla_v f(a) $$ So while $\nabla f$ is a vector, $df$ is a function (which happens to use $\nabla f$ in its definition). It just happens that the matrix representation of the function $df$ has the same components as the gradient vector, since matrix multiplication is the same as taking the dot product.

I think this also answers your question #2.

  • $\begingroup$ the gradient is a derivative which when evaluated gives a covector and yes! is a vector but in the dual space ${\mathbb R^n}^*$ $\endgroup$
    – janmarqz
    Dec 31, 2020 at 1:50
  • $\begingroup$ @janmarqz Actually, it is the differential $df$ that is a covector. The gradient is a vector, which is dual to $df$, using a Riemannian metric (which in the most basic situation is just the ordinary dot product). But it sounded like the original poster did not have the necessary background to talk about differential forms and Riemannian metrics, so I avoided this discussion. $\endgroup$
    – Nick
    Dec 31, 2020 at 2:01
  • $\begingroup$ $df_a(v) = \triangledown_v f(a)$, so why not call the total derivative the directional derivative? Unless for the total derivative $v$ is perhaps expected to be $v = k(e_1 + .. + e_n)$ vs a particular direction? $\endgroup$
    – Nick
    Dec 31, 2020 at 2:02
  • $\begingroup$ The total derivative is just the $df$ (without the $v$). It is a function, which when you plug in $v$, gives the directional derivative. $\endgroup$
    – Nick
    Dec 31, 2020 at 2:03
  • $\begingroup$ the gradient of a scalar function is a derivative and has functions as components then is not a vector neither a covector, it is a covector (a linear functional) when the component functions are evaluated at some location in the domain of the function $\endgroup$
    – janmarqz
    Dec 31, 2020 at 2:07

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