$Let $ X = x i+ y j+z k$, and $r=||X||$. Prove that $\nabla (\frac{1}{r})=-\frac{X}{r^3}.$$

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$Let $ X = x i+ y j+z k$, and $r=||X||$. Prove that $\nabla (\frac{1}{r})=-\frac{X}{r^3}.$$

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Let $ X = x i+ y j+z k$, and $r=||X||$. Prove that $\nabla (\frac{1}{r})=-\frac{X}{r^3}.$

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Multivariable Calculus Calculus

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Note that $r=\sqrt{x^2+y^2+z^2}$. Then using the chain rule we have $\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2+z^2}}$. So
\[\frac{\partial }{\partial x}(\frac{1}{r})=-\frac{1}{r^2}\frac{\partial r}{\partial x}=-\frac{1}{r^2}\frac{x}{\sqrt{x^2+y^2+z^2}}=-\frac{1}{r^3}x.\]
Similarly we can show that
\[\frac{\partial }{\partial y}(\frac{1}{r})=-\frac{1}{r^3}y \text{and} \frac{\partial }{\partial z}(\frac{1}{r})=-\frac{1}{r^3}z.\]
Hence
\[\nabla (\frac{1}{r})=(-\frac{1}{r^3}x , -\frac{1}{r^3}y , -\frac{1}{r^3}z)=-\frac{1}{r^3}X.\]

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$Let $ X = x i+ y j+z k$, and $r=||X||$. Prove that $\nabla (\frac{1}{r})=-\frac{X}{r^3}.$$

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