Gradient of a scalar to a power

Question

I was working on a problem and had a question about how gradients worked. I have the following expression:$\nabla (\rho^{-\gamma})$ and was wondering if I could express it as: \begin{gather} \nabla (\rho^{-\gamma}) = -\gamma\rho^{-\gamma-1} \nabla \rho, \end{gather} where $\rho$ is a scalar. Just doing some trial cases, I see that it works for $\nabla(\rho^2)$ with $\rho(x,y)=2xy^2$ for example. So I do not see why this would not be the case here, but how would you prove it?

Answer 1

That should work, it follows from using the chain rule for each of the derivatives in the gradient.

Let's say $\rho$ is a function of $x, y, $ and $z$. (it works completely fine for other dimensions too)

It should be clear from our differentiation rules that $\dfrac{\partial}{\partial x} \rho^{-\gamma} = -\gamma \rho^{-\gamma - 1}\dfrac{\partial\rho}{\partial x}$, and likewise for $y$ and $z$.

So, if we write out our gradient, we have $\nabla [\rho^{-\gamma}] = \begin{bmatrix}-\gamma \rho^{-\gamma - 1}\dfrac{\partial\rho}{\partial x} \\ -\gamma \rho^{-\gamma - 1}\dfrac{\partial\rho}{\partial y} \\ -\gamma \rho^{-\gamma - 1}\dfrac{\partial\rho}{\partial z}\end{bmatrix} = -\gamma \rho^{-\gamma - 1}\begin{bmatrix}\dfrac{\partial\rho}{\partial x} \\ \dfrac{\partial\rho}{\partial y} \\ \dfrac{\partial\rho}{\partial z}\end{bmatrix} = -\gamma \rho^{-\gamma - 1} \nabla \rho$.