0
$\begingroup$

Suppose $f: \mathbb{R}^n \to \mathbb{R}$ is a differentiable function. Then we can write the derivative of $f$ as a $1 \times n$ row matrix of partial derivatives of $f$ ,i,e,

$$Df=\begin{bmatrix}\frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} & \cdots & \frac{\partial f}{\partial x_n}\end{bmatrix}$$

Now suppose I wish to find the derivative of $f$ along the line $p+tv$ where $p$ is an arbitrary point and $v$ is a tangent vector and $t$ is a parameter. If I calculate the quantity $\frac{d(f(p+vt))}{dt}$, what exactly is this thing? Intuitively I think it is the rate of change of $f$ along the line $p+tv$,i.e, rate of change which occurs if we move along this line. But how is it related to the row matrix derivative above? If I substitute the coordinates by $p_i +tv_i$ in the above row matrix then how is that derivative different from this ?

$\endgroup$
1

1 Answer 1

1
$\begingroup$

Note that you want to compute:

$$(f(g(t)))'$$

Where $f$ is your function and:

$$g:\mathbb{R} \to \mathbb{R}^n$$

$$g(t)=p+vt$$

To compute $(f(g(t))'$ you can use chain rule. Note thet chain rule gives:

$$(f(g(t))'=(Df)(p+vt)v$$

where $Df$ is matrix of first partial derivatives.

$$Df(p+vt)=\begin{bmatrix}\frac{\partial f}{\partial x_1}(p+vt) & \frac{\partial f}{\partial x_2}(p+vt) & \cdots & \frac{\partial f}{\partial x_n}(p+vt)\end{bmatrix}$$

$\endgroup$
6
  • $\begingroup$ But what if I put p+tv into the function f, make it a function of t alone and then differentiate it wrt t? $\endgroup$
    – RagingBull
    Aug 5, 2015 at 16:36
  • $\begingroup$ What you are really doing is making new function $f \circ g: \mathbb{R} \to \mathbb{R}$. To differentiate $f \circ g$ you use chain rule en.wikipedia.org/wiki/Chain_rule $\endgroup$
    – agha
    Aug 5, 2015 at 16:39
  • $\begingroup$ got it! thanks a lot :) $\endgroup$
    – RagingBull
    Aug 5, 2015 at 17:08
  • $\begingroup$ ok so we have two things: one is the derivative of f at the point p+tv which is a row matrix and the other is derivative of the composite function f(p+tv).Am I right? How exactly are the two things different? I mean geometrically. I understand that the latter gives the rate of change of f along the curve p+tv. $\endgroup$
    – RagingBull
    Aug 5, 2015 at 17:16
  • 1
    $\begingroup$ Yes, you are right. One thing, derivative of $f$ is matrix, it's $Df$. But you can compute rate of change of $f$ along different vectors. If you want compute rate of change along of vector $v$ (starting at point $p$) you compute $Df(p+vt)v$ (it is matrix multiplication giving as $1 \times 1$ matrix, which is an number). $\endgroup$
    – agha
    Aug 5, 2015 at 17:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .