1. ## Basic Differentiation Rule

Hi

I recently saw this equation in a physics book:

dV . V is equivalent to
dt

d V^2
dt 2

I can see that it works but what rule of calculus tells us that when a derivative of a function is multiplied by the function itself it is equivalent to the derivative of the function squared divide by two?

I'm sure this is a very basic rule but I seem to be missing something!

2. $\frac{v^2}{2}=\frac{v.v}{2}$

Now differentiate with respect to time. (Use the chain rule.)

3. Thanks Agent Mahone.

I understand how (v.v)/2 becomes v.dv/dt using the chain rule

What I didn't understand was how to recognise the reverse. I still don't quite get it but I'll give it some thought.!

Thanks again.

4. Originally Posted by lonelyoutpost
Thanks Agent Mahone.

I understand how (v.v)/2 becomes v.dv/dt using the chain rule

What I didn't understand was how to recognise the reverse. I still don't quite get it but I'll give it some thought.!

Thanks again.
If a= b then b= a!

If d(v^2)/dt= 2v(dv/dt) then d(v^2/2)dt= v(dv/dt)
so that v(dv/dt)= d(v^2/2)/dt.

5. Thanks for your help. I think maybe I shouldn't have used the word 'equation' in my original question. I realise that the maths works, that the two sides of the equation are equal and that a=b and therefore b=a.

What I should have said from the start is...

The actual equation I saw was:

dV/dt = a

and then the author said to multiply both sides by V giving:

d/dt (V^2)/2 = av

I was confused at that point because I couldn't see why the LHS wasn't written:

(dV/dt) . V

The logical jump to reverse the chain rule just hadn't occurred to me. And it still wouldn't (although I can see why it works). I think the problem for me is that I haven't yet grasped a deep, intuitive feel for what the chain rule actually means.

Anyway, thanks all for your help. I need to give it some thought. If someone can explain the answer to my question (which I find very difficult even to word right) I'd be interested to hear from you!

6. Originally Posted by lonelyoutpost
Thanks for your help. I think maybe I shouldn't have used the word 'equation' in my original question. I realise that the maths works, that the two sides of the equation are equal and that a=b and therefore b=a.

What I should have said from the start is...

The actual equation I saw was:

dV/dt = a

and then the author said to multiply both sides by V giving:

d/dt (V^2)/2 = av

I was confused at that point because I couldn't see why the LHS wasn't written:

(dV/dt) . V

The logical jump to reverse the chain rule just hadn't occurred to me. And it still wouldn't (although I can see why it works). I think the problem for me is that I haven't yet grasped a deep, intuitive feel for what the chain rule actually means.

Anyway, thanks all for your help. I need to give it some thought. If someone can explain the answer to my question (which I find very difficult even to word right) I'd be interested to hear from you!
They left some steps out, is all.

The simplest way to approach this is to first take a look at (d/dt)v^2:
$\frac{d}{dt}(v^2) = 2v~\frac{dv}{dt}$

So
$v~\frac{dv}{dt} = \frac{1}{2} \frac{d}{dt}(v^2)$

With that said:
$a = \frac{dv}{dt}$

$av = v \frac{dv}{dt} = \frac{1}{2} \frac{d}{dt}(v^2)$

By the way, Physics and Math are "case sensitive." So the variables V and v are not the same thing. In Physics velocity and speed are usually given by the variable v. V tends to stand for a total velocity, when it's not volume or electric potential (voltage.)

-Dan

7. Hi Dan

Thanks for taking the time to reply. I take your point about case sensitivity and will try not to make the same mistake again!

You've explained it very well but it seems to me that to take the very last step in your logic requires the prior knowledge you specified in the first two lines. That's fair enough, of course, but if you don't have that snippet of knowledge how do you work it out? Is it trial and error? What I mean is there seems to be no rule which allows you to progress from left to right in that final step (it seems arbitrary) whereas going from right to left is just a simple case of differentiation using the chain rule.

I'm sure I'm making this far more complicated than it is!

Thanks again.

Matt

8. Originally Posted by lonelyoutpost
Hi Dan

Thanks for taking the time to reply. I take your point about case sensitivity and will try not to make the same mistake again!

You've explained it very well but it seems to me that to take the very last step in your logic requires the prior knowledge you specified in the first two lines. That's fair enough, of course, but if you don't have that snippet of knowledge how do you work it out? Is it trial and error? What I mean is there seems to be no rule which allows you to progress from left to right in that final step (it seems arbitrary) whereas going from right to left is just a simple case of differentiation using the chain rule.

I'm sure I'm making this far more complicated than it is!

Thanks again.

Matt
Well it helps that I've seen it before. Sometimes you have to see it to know how to do it. (That's why we make the big money here on MHF. )

-Dan