Question

Show that ∇𝑟ⁿ = 𝑛𝑟^n-2$\vec r$

Answer 1

$\triangledown r^{n}=nr^{n-2}\vec r$

$\vec r=x\hat i+y\hat j+z\hat k$

$|\vec r|^{n}=r^{n}=\sqrt [n] {x^{2}+y^{2}+z^{2}}$

$\triangledown=\frac{d}{dx}\hat i+\frac{d}{dy}\hat j+\frac{d}{dz}\hat k$

$\triangledown r^{n}=\frac{dr^{n}}{dx}\hat i+\frac{dr^{n}}{dy}\hat j+\frac{dr^{n}}{dz}\hat k$

$\frac{dr^{n}}{dx}=\frac{d(x^{2}+y^{2}+z^{2})^{\frac{n}{2}}}{dx}$

$=\frac{n}{2}(x^{2}+y^{2}+z^{2})^{{\frac{n}{2}}-1}\frac{d}{dx}(x^{2}+y^{2}+z^{2})$

$=\frac{n}{2}(x^{2}+y^{2}+z^{2})^{{\frac{n}{2}}-1}2x$

$=n(x^{2}+y^{2}+z^{2})^{{\frac{n}{2}-1}}x$

$||y$

$\frac{dr^{n}}{dy}=\frac{d(x^{2}+y^{2}+z^{2})^{\frac{n}{2}}}{dy}$

$=n(x^{2}+y^{2}+z^{2})^{\frac{n}{2}-1}y$

And;

$\frac{dr^{n}}{dz}=\frac{d(x^{2}+y^{2}+z^{2})^{\frac{n}{2}}}{dz}$

$=n(x^{2}+y^{2}+z^{2})^{\frac{n}{2}-1}z$

Hence;

$\triangledown r^{n}=n(x^{2}+y^{2}+z^{2})^{\frac{n}{2}-1}(x\hat i+y\hat j+z\hat k)\newline \triangledown r^{n}=n(x^{2}+y^{2}+z^{2})^{\frac{n}{2}-1}\vec r$

$\triangledown r^{n}=n(x^{2}+y^{2}+z^{2})^{\frac{n-2}{2}}\vec r$

Thus,

$\triangledown r^{n}=nr^{n-2}\vec r$