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.