Problem:

For each of the following functions, find the derivative vector $\displaystyle \nabla f(x)$ for those points $\displaystyle x \in \mathbb{R}^n$ where it is defined:

a. $\displaystyle f(x)=e^{\| x \| ^2}$

c.$\displaystyle f(x) = \frac{1}{\| x \|^2}$

================================

a. I know that $\displaystyle \| x \| = \sqrt{x_{1}^2+x_{2}^2+ ... + x_{n}^2}$. As a result, then $\displaystyle \| x \|^2 = x_{1}^2+x_{2}^2+ ... + x_{n}^2 $

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

By taking the partial derivatives, then

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

....

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

So, the gradient would be

$\displaystyle \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) = $

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

However, I have a hard time understanding the gradient vector/derivative vector. I'm slightly confused because I thought $\displaystyle \nabla f(x)$ 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.