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I've got confused in something which should not be too confusing. But...

If we have some function $f(r)$, where $r=\sqrt{x^2+y^2+z^2}$, then what is $f'(r) = \frac{df}{dr}$?

I thought it would be this: $\frac{df}{dr} = \frac{∂r}{∂x} + \frac{∂r}{∂x} + \frac{∂r}{∂x}$, but something is not right here. What is confusing me so much is the absense of the part differentiating the variable $r$ itself.

Say, if I had $f(r) = r^2$, I would say $f'(r) = 2r(\frac{∂r}{∂x} + \frac{∂r}{∂x} + \frac{∂r}{∂x})$. But $2r$ is $\frac{df}{dr}$... Please help :)

sequence
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  • I can try with this notation $f'(r) = r' (r_x + r_y + r_z)$, but then how can it be expressed in Leibniz notation? – sequence Sep 20 '15 at 00:53
  • If you have $f(r) = r^2$, then $f'(r) = 2r$. If you have $g(x,y,z) = f(\sqrt{x^2+y^2+z^2})$, then the derivative of $g$ is a map $\mathbb{R}^3 \to \mathbb{R}$. – copper.hat Sep 20 '15 at 01:05
  • What will $g'$ look like? – sequence Sep 20 '15 at 01:16
  • For example, ${\partial g(x,y,z) \over \partial x} = f'(\sqrt{x^2+y^2+z^2}){x \over \sqrt{x^2+y^2+z^2}}$. – copper.hat Sep 20 '15 at 01:19
  • Some abuse of notation is reasonable in the context of established notation and economy, but much is sloppy and almost guaranteed to introduce confusion. – copper.hat Sep 20 '15 at 02:39

1 Answers1

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Think of $r$ as a function of three variables:

$$r(x,y,z)=\sqrt{x^2+y^2+z^2}$$

$r$ is therefore a function from $\mathbb{R}^3$ to $[0,\infty)$.

Now you have a function $f$ defined on $[0,\infty)$, whose range is $\mathbb{R}$, say, and you are looking at the composition $f\circ r$. This function starts in $\mathbb{R}^3$, and ends up in the range of $f$. By a common abuse of notation, you choose to denote this function by $f(r)$, but what it really means is the composition $f\circ r$. If you are going to differentiate it as a function from $\mathbb{R}^3$ to $\mathbb{R}$, you will be relying on the chain rule, which tells you that: $${\partial (f\circ r)\over\partial x}(x_0,y_0,z_0)=f'(r(x_0,y_0,z_0))\cdot {\partial r\over \partial x}(x_0,y_0,z_0)$$ and by further abuse of notation, this is often denoted as $${df\over dr}{\partial r\over \partial x}\enspace{\rm{or}}\enspace f'(r){\partial r\over \partial x}$$

In a similar way you obtain the other two partial derivatives. The result is a vector of three components, known as the gradient of $f$, which can be written as $f'(r)\nabla r$, where $\nabla r$ is the gradient of $r$, and $f'(r)$, is a number, both evaluated at the point $(x_0,y_0,z_0)$. So the derivative of $f\circ r$ at the point $(x_0,y_0,z_0)$ is the vector:

$$f'(r(x_0,y_0,z_0))\cdot({\partial r\over \partial x}(x_0,y_0,z_0),{\partial r\over \partial y}(x_0,y_0,z_0),{\partial r\over \partial z}(x_0,y_0,z_0))$$

uniquesolution
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  • So it appear to be that $\vec{\nabla} f(r)=f′(r) * 1/r * (x,y,z) = f'(r) \vec{r}/{r}$, if I'm right. The abuse of notation gets me confused in applied math... – sequence Sep 20 '15 at 01:42
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    Yes, in this particular case the gradient of $r$ at any vector is just $x/||x||$, which is $x/r(x)$. I don't like the notation $\nabla f(r)$, too abusive, if you ask me. Anyway, $f'(r)(x,y,z)/r(x,y,z)$ is indeed the gradient of $f\circ r$. – uniquesolution Sep 20 '15 at 01:44