Let $g$ be a function of $r$, $r=\left\|{X}\right\|$, where $f(X)=g(r)$. Show that
$(\frac{dg}{dr})^2=(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2+(\frac{\partial f}{\partial z})^2$
I need some keys for solve this. All I have noted is that $f$ is a function that depends on $x,y,z$ and we know that $g(r)=f(X)=f(x,y,z)$. So,
$\frac{dg}{dr}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}$
I know that something is wrong with the right side of the last equation because I have nothing to say that I am taking the derivative with respect to r, i.e. the length of $X$, but I don´t know how to put that. Also, I was wondering if the squares in the first equation come from the definition of the norm of $X $
$\left\|{X}\right\|^2=x^2+y^2+z^2$
What should I do next?I have to take the derivative of each one of the arguments of $f$ with respect to the norm? But the norm again depends on $x,y,z$
Help, please!!!!! I´m lost
Note that $g(r)=g(\sqrt{x^2+y^2+z^2})=f(X)$. So, by the chain rule $$\frac{\partial f}{\partial x}=\frac{dg}{dr}\frac{\partial r}{\partial x}=\frac{dg}{dr}\frac{x}{\sqrt{x^2+y^2+z^2}}$$ and so on.
Then, when you square and sum the three derivatives of $f$ you get $$\left(\frac{dg}{dr}\right)^2\frac{x^2}{r^2}+\left(\frac{dg}{dr}\right)^2\frac{y^2}{r^2}+\left(\frac{dg}{dr}\right)^2\frac{z^2}{r^2}=\frac{1}{r^2}\left(\frac{dg}{dr}\right)^2r^2=\left(\frac{dg}{dr}\right)^2$$