VC3.3

i). $\nabla f(r)=[i\frac{\partial}{\partial x}+j\frac{\partial}{\partial y}+k\frac{\partial}{\partial z}]f(r)\,$

= $i\frac{\partial}{\partial x}f(r)+j\frac{\partial}{\partial y}f(r)+k\frac{\partial}{\partial z}f(r)\,$

= $if'(r)\frac{\partial r}{\partial x}+jf'(r)\frac{\partial r}{\partial y}+kf'(r)\frac{\partial r}{\partial z}\,$

= $f'(r)[i\frac{\partial r}{\partial x}+j\frac{\partial r}{\partial y}+k\frac{\partial r}{\partial z}]\,$

= $f'(r)\nabla r\,$

ii). we have $\nabla r=i\frac{\partial r}{\partial x}+j\frac{\partial r}{\partial y}+k\frac{\partial r}{\partial z}\,$

Since $r=xi+yj+zk\,$ So, $r^2=|r|^2=x^2+y^2+z^2\,$

Differentiating the above,we get

$2r\frac{\partial r}{\partial x}=2x,\frac{\partial r}{\partial x}=\frac{x}{r}\,$

Similarly, $\frac{\partial r}{\partial y}=\frac{y}{r},\frac{\partial r}{\partial z}=\frac{z}{r}\,$

Therefore,by using the above, $\nabla r=\frac{x}{r}i+\frac{y}{r}j+\frac{z}{r}k\,$

= $\frac{1}{r}(xi+yj+zk)=\frac{1}{r}r\,$