• Aug 21st 2008, 03:01 PM
Paperwings
Problem:
For each of the following functions, find the derivative vector $\nabla f(x)$ for those points $x \in \mathbb{R}^n$ where it is defined:
a. $f(x)=e^{\| x \| ^2}$
c. $f(x) = \frac{1}{\| x \|^2}$
================================
a. I know that $\| x \| = \sqrt{x_{1}^2+x_{2}^2+ ... + x_{n}^2}$. As a result, then $\| x \|^2 = x_{1}^2+x_{2}^2+ ... + x_{n}^2$

So, if $f(x)=e^{\| x \| ^2} \implies f(x) = e^{x_{1}^2+x_{2}^2+ ... + x_{n}^2}$

By taking the partial derivatives, then

$\frac{\partial f}{\partial x_{1}}(x)= 2x_{1}e^{x_{1}^2+x_{2}^2+ ... + x_{n}^2}$

....

$\frac{\partial f}{\partial x_{n}}(x)= 2x_{n}e^{x_{1}^2+x_{2}^2+ ... + x_{n}^2}$

$\nabla f(x) = \left( \frac{\partial f}{\partial x_{1}}(x), \frac{\partial f}{\partial x_{2}}(x), ....,\frac{\partial f}{\partial x_{n}}(x) \right) =$

$\left( 2x_{1}e^{x_{1}^2+x_{2}^2+ ... + x_{n}^2}, ..., 2x_{n}e^{x_{1}^2+x_{2}^2+ ... + x_{n}^2}\right)$

However, I have a hard time understanding the gradient vector/derivative vector. I'm slightly confused because I thought $\nabla f(x)$ was called the gradient and not a vector. If I can just understand this, then I can do (c) as well.

• Aug 21st 2008, 03:57 PM
Spec
Quote:

Originally Posted by Paperwings
However, I have a hard time understanding the gradient vector/derivative vector. I'm slightly confused because I thought $\nabla f(x)$ was called the gradient and not a vector. If I can just understand this, then I can do (c) as well.

It is called the gradient, but it's also a vector. In fact, the gradient is the normal vector to the point $\nabla f(\overline{x})$, and it also points in the direction that the function is growing the fastest at that point.
• Aug 22nd 2008, 03:55 AM
Paperwings
Thank you Spec. I came to realized that my book referred to the gradient as the gradient vector or derivative vector.
• Aug 22nd 2008, 05:22 AM
wingless
You can remember the definition of del.

For $\mathbb{R}^3$,
$\nabla = \frac{\partial}{\partial x}\mathbf{i} + \frac{\partial}{\partial y}\mathbf{j} + \frac{\partial}{\partial z}\mathbf{k}$

For $\mathbb{R}^n$,
$\nabla = \left ( \frac{\partial}{\partial x_1}, \frac{\partial}{\partial x_2} ,\frac{\partial}{\partial x_3}, \dots , \frac{\partial}{\partial x_n} \right ) = \sum_{i=1}^{n} \frac{\partial}{\partial x_i}\vec{e}_i$
(that $\vec{e}_i$ is the standart basis of the space)

So as you see here, del is taken as a vector.

Now we can derive the operators (for $\mathbb{R}^3$).

Let $f(x,y,z)$ be a scalar function.
$\text{grad} f = \nabla f = \left (\frac{\partial}{\partial x}\mathbf{i} + \frac{\partial}{\partial y}\mathbf{j} + \frac{\partial}{\partial z}\mathbf{k}\right ) f = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} + \frac{\partial f}{\partial z}\mathbf{k}$

What we did here was multiplying a vector (del) by a scalar (f).

Let F be a vector function such that $F(x,y,z) = M\mathbf{i} + N \mathbf{j} + P \mathbf{k}$.

$\text{div} F = \nabla \cdot F = \left (\frac{\partial}{\partial x}\mathbf{i} + \frac{\partial}{\partial y}\mathbf{j} + \frac{\partial}{\partial z}\mathbf{k}\right )\cdot \left ( M\mathbf{i} + N \mathbf{j} + P \mathbf{k} \right ) = \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} + \frac{\partial P}{\partial z}$

We just applied dot product here.

And curl,
$\text{curl} F = \nabla \times F = \left (\frac{\partial}{\partial x}\mathbf{i} + \frac{\partial}{\partial y}\mathbf{j} + \frac{\partial}{\partial z}\mathbf{k}\right )\times \left ( M\mathbf{i} + N \mathbf{j} + P \mathbf{k} \right ) = \left|
\begin{array}{ccc}
i & j & k \\
\tfrac{\partial }{\partial x} & \tfrac{\partial }{\partial y} & \tfrac{\partial }{\partial z} \\
M & N & P
\end{array}
\right|$