I'm having a bit of difficulty with this problem.
Show that gradient(f^r) = r*f^(r-1)*gradient(f)
Not even sure where to begin with this, thanks!
Let us assume $\displaystyle f(x_1,...,x_n)$ is a differenciable function.
Thus,
$\displaystyle \nabla f=\left< \frac{\partial f}{\partial x_1},..., \frac{\partial f}{\partial x_n} \right>$
Thus,
$\displaystyle \nabla f^r = \left< \frac{\partial f^r}{\partial x_1},...,\frac{\partial f^r}{\partial x_n} \right>$.
But, by the chain rule,
$\displaystyle \frac{\partial f^r}{\partial x_i}=rf^{r-1}\frac{\partial f}{\partial x_i}$ for $\displaystyle i=1,2,...,n$.
Thus,
$\displaystyle \nabla f^r = \left< rf^{r-1} \frac{\partial f}{\partial x_1},...,rf^{r-1}\frac{\partial f}{\partial x_n} \right>=rf^{r-1}\left< \frac{\partial f}{\partial x_1},...,\frac{\partial f}{\partial x_n} \right>=rf^{r-1}\nabla f$