How to prove this statement about the derivative of a function

[oliverkahn]

Homework Statement

To prove: $\dfrac{d[f(r)]}{dr}=$ constant

Relevant Equations

Given (1): $r=\sqrt{a^2 + p^2 - 2 ap \cos \theta}$

where $a,p,\theta$ are independent

Given (2): $\dfrac{d[f(a+p)]}{dp}=\dfrac{d[f(a-p)]}{dp}$

My try:

$\begin{align} \dfrac{d {r^2}}{d r} \dfrac{\partial r}{\partial p} = \dfrac{\partial {r^2}}{\partial p} \tag1\\ \dfrac{\partial r}{\partial p} = \dfrac{\partial {r^2}}{\partial p} \dfrac{1}{\dfrac{d r^2}{d r}}=\dfrac{p-a\cos\theta}{r} \tag2\\ \end{align}$

By chain rule:

$\dfrac{\partial [f(r)]}{\partial p}=\dfrac{d [f(r)]}{d r} \dfrac{\partial r}{\partial p} = \dfrac{d [f(r)]}{d r} \dfrac{p-a\cos\theta}{r}$

If it is the right approach, please complete the proof.

I do not even know if it is the right approach.

Thanks in advance $\forall$ help.

 

[Paul Colby]

Suppose $f(r) = r^7$.

$\frac{d f(r)}{dr} = 7r^6 \ne \text{constant}$

am I good so far?

 

[oliverkahn]

This means we have to show $f(r)=r+C$. Or am I missing anything?

 

[member 587159]

Read your own question and imagine you are someone else reading it. Does that person know what $f$ is?

If not, you have some editing to do.

 

[oliverkahn]

Oh! Ok... Let me edit it.

 

[Mark44]

To prove: $\dfrac{d[f(r)]}{dr}=$ constant

This means we have to show $f(r)=r+C$. Or am I missing anything?

What you're trying to prove amounts to solving a simple differential equation.
$\dfrac{d[f(r)]}{dr}= k \Rightarrow d~f(r) = k~dr \Rightarrow \int d~f(r) = \int k~dr$
Can you take it from here?

 

[oliverkahn]

Actually I am trying to understand the paragraphs in the following article:

My teacher has shown the proof to me only until the fourth last equation, i.e. $0= f'(a+p)-f'(a-p)$

From here I have to get $f'(r)=C$. Any clue?

 

[Paul Colby]

Since $a$ and $p$ are independent, let $x=a+p$ and $y=a-p$. $x$ and $y$ are also independent. So $f'(x) = f'(y)$ for every $x$ and $y$. Pretty much the definition of constant.

 

[archaic]

I also have a question about this:
$$r^2=a^2-2ap\cos\theta+p^2\Rightarrow 2rdr=2ap\sin\theta$$
Why can he consider the differential with respect to $r$ as equal to the differential with respect to $\theta$?
Never mind, chain rule.