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Given a function $V(x)$: $$\displaystyle V(x) = \frac {1}{||x||}$$ What is the gradient, $\nabla V$, of $V$?

The result I saw is: $\displaystyle -\frac {x}{||x||^3}$.

How do I get this?

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    $\begingroup$ You should complete the question showing what you have tried. $\endgroup$
    – user
    Aug 23 '21 at 21:29
  • $\begingroup$ More explicitly, $V(x_1,\dots, x_n)=\frac{1}{\sqrt{(x_1)^2+\cdots +(x_n)^2}}$. I'm sure using the quotient rule and standard rules of differentiation, you can calculate $\frac{\partial V}{\partial x_i}$ $\endgroup$
    – peek-a-boo
    Aug 23 '21 at 21:31
  • $\begingroup$ Hint, how do you calculate the magnitude of a vector given it's coordinates? $\endgroup$ Aug 23 '21 at 21:33
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Using the definition of vector norm $||x||=\sqrt{\sum\limits_{i}{x_i^2}}$, we have, $\frac{\partial V}{\partial x_i}=-\frac{1}{||x||^2}.\frac{1}{2||x||}.2x_i$, by chain rule. Hence, $\nabla V=\left[-\frac{x_1}{||x||^3}, -\frac{x_2}{||x||^3}, \ldots\right]^T=-\frac{x}{||x||^3}$.

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Let's first rename your function like so: $$ f(\vec{\bf{v}})=\frac{1}{||\vec{\bf{v}}||} $$ We can then name each component of your vector $\vec{\bf{v}}_x$ and $\vec{\bf{v}}_y$, and have them as the individual parameters of the function and given that $||\vec{\bf{v}}||=\sqrt{\left(\vec{\bf{v}}_x\right)^2+\left(\vec{\bf{v}}_y\right)^2}$: $$ \begin{align} & f(\vec{\bf{v}}_x,\vec{\bf{v}}_y)=\frac{1}{||\vec{\bf{v}}||} \\ \implies & f(\vec{\bf{v}}_x,\vec{\bf{v}}_y)=\frac{1}{\sqrt{\left(\vec{\bf{v}}_x\right)^2+\left(\vec{\bf{v}}_y\right)^2}} \end{align} $$ Now we can differentiate this with respect to $\vec{\bf{v}}_x$ or $\vec{\bf{v}}_y$ like so: $$ \begin{align} &f(\vec{\bf{v}}_x,\vec{\bf{v}}_y)=\left(\left(\vec{\bf{v}}_x\right)^2+\left(\vec{\bf{v}}_y\right)^2\right)^\frac{-1}{2} \\ &\frac{\partial f}{\partial \vec{\bf{v}}_x} = \frac{-1}{2}\left(\left(\vec{\bf{v}}_x\right)^2+\left(\vec{\bf{v}}_y\right)^2\right)^\frac{-3}{2}\cdot\frac{\partial \left(\left(\vec{\bf{v}}_x\right)^2+\left(\vec{\bf{v}}_y\right)^2\right)}{\partial \vec{\bf{v}}_x} \\ &\frac{\partial f}{\partial \vec{\bf{v}}_x} = \frac{-1}{2}\left(\left(\vec{\bf{v}}_x\right)^2+\left(\vec{\bf{v}}_y\right)^2\right)^\frac{-3}{2}\cdot 2 \vec{\bf{v}}_x \\ &\frac{\partial f}{\partial \vec{\bf{v}}_x} = \frac{-2\vec{\bf{v}}_x}{2}\frac{1}{\left(\sqrt{\left(\vec{\bf{v}}_x\right)^2+\left(\vec{\bf{v}}_y\right)^2}\right)^3} \\ &\frac{\partial f}{\partial \vec{\bf{v}}_x} = \boxed{\frac{-\vec{\bf{v}}_x}{\left(\sqrt{\left(\vec{\bf{v}}_x\right)^2+\left(\vec{\bf{v}}_y\right)^2}\right)^3}} \end{align} $$ Which simplifies to: $$ \frac{\partial f}{\partial \vec{\bf{v}}_x} = \boxed{\frac{-\vec{\bf{v}}_x}{||\vec{\bf{v}}||^3}} $$ And the same for differentiating with respect to $\vec{\bf{v}}_y$. So the gradient of your function is: $$ \begin{align} \nabla f &= \begin{bmatrix} \frac{-\vec{\bf{v}}_x}{||\vec{\bf{v}}||^3} \\ \frac{-\vec{\bf{v}}_y}{||\vec{\bf{v}}||^3} \end{bmatrix} \\ &=\frac{-1}{||\vec{\bf{v}}||^3}\begin{bmatrix} \vec{\bf{v}}_x \\ \vec{\bf{v}}_y \end{bmatrix} \\ &=\frac{-1}{||\vec{\bf{v}}||^3}\vec{\bf{v}} \\ &=\boxed{-\frac{\vec{\bf{v}}}{||\vec{\bf{v}}||^3}} \end{align} $$

I hope this helped!

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