91影视

Given the vector \(\mathbf{r}=\langle x, y, z\rangle\), we can define the function \(f(\mathbf{r})\) as \(f(\mathbf{r})=\frac{1}{|\mathbf{r}|^p}\), where \(|\mathbf{r}|=\sqrt{x^2+y^2+z^2}\).

The gradient of the function is a vector given by the partial derivatives of the function with respect to each coordinate: $$ \nabla f(\mathbf{r}) = \left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle $$

We can use the Chain Rule to determine the partial derivatives: $$ \frac{\partial f}{\partial x} = \frac{\mathrm{d}f}{\mathrm{d}|\mathbf{r}|} \frac{\partial|\mathbf{r}|}{\partial x} $$ First, we will find the derivative of \(f\) with respect to \(|\mathbf{r}|\): $$ \frac{\mathrm{d}f}{\mathrm{d}|\mathbf{r}|} = \frac{\mathrm{d}}{\mathrm{d}|\mathbf{r}|}\left(\frac{1}{|\mathbf{r}|^p}\right) = \frac{-p}{|\mathbf{r}|^{p+1}} $$ For the derivative of \(|\mathbf{r}|\) with respect to each coordinate, we can differentiate the expression for \(|\mathbf{r}|=\sqrt{x^2+y^2+z^2}\): $$ \frac{\partial |\mathbf{r}|}{\partial x} = \frac{x}{|\mathbf{r}|} $$ Using the Chain Rule, we obtain the partial derivative of \(f\) with respect to \(x\): $$ \frac{\partial f}{\partial x} = \frac{-p}{|\mathbf{r}|^{p+1}} \cdot \frac{x}{|\mathbf{r}|} = \frac{-px}{|\mathbf{r}|^{p+2}} $$ Similarly, we find the partial derivatives with respect to \(y\) and \(z\): $$ \frac{\partial f}{\partial y} = \frac{-py}{|\mathbf{r}|^{p+2}} $$ $$ \frac{\partial f}{\partial z} = \frac{-pz}{|\mathbf{r}|^{p+2}} $$

Now that we have all the partial derivatives, we can form the gradient of \(f(\mathbf{r})\): $$ \nabla f(\mathbf{r}) = \left\langle \frac{-px}{|\mathbf{r}|^{p+2}}, \frac{-py}{|\mathbf{r}|^{p+2}}, \frac{-pz}{|\mathbf{r}|^{p+2}} \right\rangle = \frac{-p\mathbf{r}}{|\mathbf{r}|^{p+2}} $$ This completes the proof that \(\nabla\left(\frac{1}{|\mathbf{r}|^p}\right) = \frac{-p\mathbf{r}}{|\mathbf{r}|^{p+2}}\).