Why is $\nabla U(r) = \frac{dU(r)}{dr} \nabla r$? Does anyone have a proof for the equation:
$$\nabla U(r) = \frac{dU(r)}{dr} \nabla r$$
Where $r=|{\bf r}|$ is the distance and $U(r)$ is a potential for a central force.
This is from page 13 of "Mechanics From Newton's Laws to Deterministic Chaos" by Florian Scheck, edition 4.
 A: The LHS expands to
$\nabla U(r) = \left [ \frac{\partial U(r)}{\partial x}, \frac{\partial U(r)}{\partial y}, \frac{\partial U(r)}{\partial z} \right ]^T$
The RHS expands to
$\frac{dU(r)}{dr} \nabla{r} = \frac{dU(r)}{dr} \left [\frac{\partial r}{\partial x}, \frac{\partial r}{\partial y}, \frac{\partial r}{\partial z} \right ]^T$ 
Assuming, by some theorem unknown to me,
$\frac{dU(r)}{dr} \frac{\partial r}{\partial x} = \frac{\partial U(r)}{\partial x} $,
then the proof is obvious.
anyone know what that theorem is?
A: It is an application of the total derivative. If partial derivatives exist, the total derivative of a scalar function $f$ of many (i.e. more than one) variables along a curve, $x:t\rightarrow x(t)$, is $$ \frac{df}{dt} = \sum_i \frac{\partial f}{\partial x^i}\frac{dx^i}{dt} $$
It is easily seen that this is true, because the change in $f$ for a small change in $t$ is simply the sum of the changes due to the change in each direction, $x^i$, while keeping the other $x^j$ constant. If you want a formal rigorous proof from first principles in analysis, you can use induction. Starting in one dimension you have simply the chain rule, then prove from first principles that if it is true in $n$ dimensions, then it is true in $n+1$ dimensions. I have given a proof in The Mathematics of Gravity and Quanta, but physicists are not usually interested in this level of rigour for intuitively clear results, and it is a routine proof for a mathematician.
A: $$U=U(r(x,y,z))$$
$$\frac{\partial U}{\partial x}=\frac{dU}{dr}\frac{\partial r}{\partial x}$$
$\Rightarrow$
$$\vec{\nabla} (U)=\begin{bmatrix}
   \frac{\partial U}{\partial x} \\
   \frac{\partial U}{\partial y} \\
   \frac{\partial U}{\partial z} \\
 \end{bmatrix}=
\begin{bmatrix}
   \frac{dU}{dr}\frac{\partial r}{\partial x} \\
  \frac{dU}{dr}\frac{\partial r}{\partial y} \\
   \frac{dU}{dr}\frac{\partial r}{\partial z} \\
 \end{bmatrix}=\frac{dU}{dr}\vec{\nabla}(r)\quad \surd$$ 
Example
$$U(r)=\frac{\mu}{r}\quad,
\frac{dU}{dr}=-\frac{\mu}{r^2}$$
with $\vec{r}=[x,y,z]^T\quad ,|\vec{r}|=\sqrt{\vec{r}\cdot\vec{r}}$
thus:
$$\frac{dU}{dr}\vec{\nabla}(|\vec{r}|)=
 \left[ \begin {array}{c} -{\frac {\mu\,x}{{r}^{2}\sqrt {{x}^{2}+{y}^{
2}+{z}^{2}}}}\\ -{\frac {\mu\,y}{{r}^{2}\sqrt {{x}^{
2}+{y}^{2}+{z}^{2}}}}\\ -{\frac {\mu\,z}{{r}^{2}
\sqrt {{x}^{2}+{y}^{2}+{z}^{2}}}}\end {array} \right] 
$$
and with ${r}=\sqrt{x^2+y^2+z^2}$ you get:
$$ \frac{dU}{dr}\vec{\nabla}(|\vec{r}|)=\left[ \begin {array}{c} -{\frac {\mu\,x}{ \left( {x}^{2}+{y}^{2}+{z}
^{2} \right) ^{3/2}}}\\-{\frac {\mu\,y}{ \left( {x}
^{2}+{y}^{2}+{z}^{2} \right) ^{3/2}}}\\ -{\frac {\mu
\,z}{ \left( {x}^{2}+{y}^{2}+{z}^{2} \right) ^{3/2}}}\end {array}
 \right] 
\tag 1$$
with $U=\frac{\mu}{\sqrt{x^2+y^2+z^2}}$
thus
$$\vec{\nabla}U(x,y,z)= \left[ \begin {array}{c} -{\frac {\mu\,x}{ \left( {x}^{2}+{y}^{2}+{z}
^{2} \right) ^{3/2}}}\\ -{\frac {\mu\,y}{ \left( {x}
^{2}+{y}^{2}+{z}^{2} \right) ^{3/2}}}\\ -{\frac {\mu
\,z}{ \left( {x}^{2}+{y}^{2}+{z}^{2} \right) ^{3/2}}}\end {array}
 \right] 
\tag 2$$
equation (1) and (2) provides the same results 
