From This article I understand that if I have a function defined using parametric equations like this $$ (1) \quad \quad \quad{{\mathbf{r}\left( t \right) = f\left( t \right)\mathbf{i} + g\left( t \right)\mathbf{j} + h\left( t \right)\mathbf{k}}\;\;\text{ or }\;\;}\kern0pt{\mathbf{r}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right),h\left( t \right)} \right\rangle } $$

Then it's derivative is the derivative of each function that generates each coordinate. $$ (2) \quad \quad \quad {\mathbf{r}^\prime\left( t \right) = \left\langle {f^\prime\left( t \right),g^\prime\left( t \right),h^\prime\left( t \right)} \right\rangle.} $$

However if the same function is represented in a non parametric form (random example below)

$$ (3) \quad \quad \quad r( x,y)= x+y $$ It seems there is no such thing as "the derivative", I only have partial derivatives, directional derivatives and the gradient.

What is the relationship between "the" derivative as defined in $(2)$ and the gradient or the partial/directional derivatives? Are they equivalent in some way?

  • $\begingroup$ Can you explain what did mean the "same function"? Because the first is an one-parameter vector-valued function and the second is a two-parameter real-valued function. $\endgroup$ – DiegoMath Mar 24 at 0:49
  • $\begingroup$ By other hand we can consider the set level of a two variable function as the image of a vector valued function, and the gradient is normal to such curve. Thus the derivative of the curve is tangent to the level set. $\endgroup$ – DiegoMath Mar 24 at 0:51
  • $\begingroup$ Sorry maybe im missing something. I mean that they draw the same surfice. On my calculus courses they are introduced without much explaination in the fact that they are or arent the same kind of object. They are mostly introduced by the picture you get when you plot them. $\endgroup$ – Joaquin Brandan Mar 24 at 0:52
  • $\begingroup$ As I understand it one function can be expressed in parametric or non parametric form, that's what I try to convey in the difference between $(1)$ and $(2)$ $\endgroup$ – Joaquin Brandan Mar 24 at 0:57

The parametrically defined function $$r(t) = \begin{pmatrix}f(t)\\g(t)\\h(t)\end{pmatrix}$$ is a function from the real number line $\mathbb R$ to the three dimensional space $\mathbb R^3$. Thus, we can take the standard one-dimensional derivative. The relationship to the directional derivative is that a direction is already chosen by the parametrisation. The directional derivative of a function $f:\mathbb R^n \to\mathbb R^m$ along a direction $v\in\mathbb R^n$ at a point $x_0$ is $$ \partial_vf(x_0) = \lim_{h\to 0} \frac{f(x_0+hv)-f(x_0)}{h}. $$ Since the parametrisation is a one-dimensional function, the only possible directions are multiples of $-1$ and $1$. Taking the directional derivative along these directions actually is equivalent to the left-sided and right-sided derivative. If $f$ is differentiable they both agree. For $r$ this is $$ r'(t) = \begin{pmatrix}f'(t)\\g'(t)\\h'(t)\end{pmatrix}. $$

The partial derivatives are actually a special case of directional derivatives. They are the directional derivatives along the coordinate axes. That is to say, for your example $g(x,y) = x+y$, the partial derivatives are $$\frac{\partial}{\partial x}g(x,y) = y\quad\text{and}\quad\frac{\partial}{\partial y}g(x,y)=x.$$

The one-dimensional derivative can be generalised for functions $f\colon\mathbb R^n\to\mathbb R^m$ to the so-called total derivative. For $m=1$ this actually is the transpose of the gradient. The function $f$ is called total differntiable in $x_0$, iff $$\lim_{x\to x_0}\frac{\lVert f(x)-f(x_0) - Df(x_0)(x-x_0)\rVert}{\lVert x-x_0\rVert},$$ where $Df(x_0)\colon \mathbb R^n\to\mathbb R^m$ is a linear map, exists. Then we call $Df(x_0)$ the total derivative of $f$ at $x_0$. Notice, the similarity to the definition of the one-dimensional derivative. If all partial derivatives exists and are continuous, then the total derivative can be expressed as $$ Df(x_0) = \biggl(\frac{\partial f}{\partial x_1},\dots, \frac{\partial f}{\partial x_n}\biggr)\in\mathbb R^{m\times n}\,. $$ Multiplying the total derivative $Df(x_0)$ with a direction gives the directional derivative.

| cite | improve this answer | |
  • $\begingroup$ Thanks for your answer. Why "a direction is already chosen by the parameterisation." ? Most of the answer I understand, the answer to my question is in your first paragraph but im not quite grasping what you say. $\endgroup$ – Joaquin Brandan Mar 24 at 1:06
  • 1
    $\begingroup$ @JoaquinBrandan I have edited the question. Let me please know if it is clearer now. $\endgroup$ – EuklidAlexandria Mar 24 at 1:59
  • 1
    $\begingroup$ @JoaquinBrandan You're welcome! Feel free to edit your question. $\endgroup$ – EuklidAlexandria Mar 24 at 2:32
  • 1
    $\begingroup$ @JoaquinBrandan Yes, you are right. Essentially, the total derivative (for real-valued functions it is the gradient) generalises this for $m>1$. $\endgroup$ – EuklidAlexandria Mar 24 at 3:14
  • 1
    $\begingroup$ @JoaquinBrandan This is only possible, if you can find a parametrisation depending on one parameter $t$. Else it is not well-defined. Better take the total derivative. In fact, the derivative you wrote is the directional derivative along $(1,1,1)^\top$. $\endgroup$ – EuklidAlexandria Mar 24 at 4:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.