# Quick derivative question

1. Oct 18, 2012

### Airsteve0

1. The problem statement, all variables and given/known data
I just have a quick question concerning the derivative of both sides of the equation shown below.

2. Relevant equations
Starting with $e^\frac{t}{m}=(\frac{v}{u})^2$

3. The attempt at a solution
Would it be correct that: $dt=(2m)(\frac{dv}{v}-\frac{du}{u})$

2. Oct 18, 2012

### Zondrina

Wait... so you have a multivariate, implicit function right? Where you want to treat t, m, v and u all as variables.

Are you trying to find $\frac{∂}{∂t}$ of both sides?

3. Oct 18, 2012

### Airsteve0

I am trying to isolate for dt so that I can use it in another equation

4. Oct 18, 2012

### Zondrina

So you want to get t to be an explicit function in terms of m, v and u and then find dt with respect to what?

5. Oct 18, 2012

### Airsteve0

v and u

6. Oct 18, 2012

### Zondrina

Okay, so first off, what is t in terms of m, v and u ( Hint : Use ln to simplify this nicely ).

Then find $\frac{∂t}{∂v}$ and $\frac{∂t}{∂u}$

7. Oct 18, 2012

### Airsteve0

well if t=2m(ln(v)-ln(u)), then I would conclude that dt=2m(dv/v) or dt=-2m(du/u). However, does this mean that dt cannot be 2m(dv/v-du/u) ?

8. Oct 18, 2012

### SammyS

Staff Emeritus
If u and v are functions of t, then yes, you are correct.

9. Oct 18, 2012

### Airsteve0

This is something that was shown in one of my classes but I am unsure of the steps to show it? Would you mind showing me just from the line t=2m(ln(v)-ln(u))?

10. Oct 18, 2012

### Zondrina

I was under the impression that you meant u, v and m were variables, but if u and v are functions of t, then you have to remember to use the chain rule here.

11. Oct 18, 2012

### Airsteve0

my apologies for not being clear, I think I have things figured out though, thanks!