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Math Help - [SOLVED] Derivative Vectors/Gradient

  1. #1
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    [SOLVED] Derivative Vectors/Gradient

    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}

    So, the gradient would be

    \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.

    Thank you for your time.
    Last edited by Paperwings; August 22nd 2008 at 04:48 AM.
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  2. #2
    Senior Member Spec's Avatar
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    Quote Originally Posted by Paperwings View Post
    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.
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  3. #3
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    Thank you Spec. I came to realized that my book referred to the gradient as the gradient vector or derivative vector.
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  4. #4
    Super Member wingless's Avatar
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    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|<br />
\begin{array}{ccc}<br />
 i & j & k \\<br />
 \tfrac{\partial }{\partial  x} & \tfrac{\partial }{\partial  y} & \tfrac{\partial }{\partial  z} \\<br />
 M & N & P<br />
\end{array}<br />
\right|
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