Derivative of Exponential Equation - Quick Question

  • Thread starter Thread starter Airsteve0
  • Start date Start date
  • Tags Tags
    Derivative
Airsteve0
Messages
80
Reaction score
0

Homework Statement


I just have a quick question concerning the derivative of both sides of the equation shown below.

Homework Equations


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

The Attempt at a Solution


Would it be correct that: dt=(2m)(\frac{dv}{v}-\frac{du}{u})
 
Physics news on Phys.org
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?
 
I am trying to isolate for dt so that I can use it in another equation
 
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?
 
v and u
 
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}
 
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) ?
 
Airsteve0 said:

Homework Statement


I just have a quick question concerning the derivative of both sides of the equation shown below.

Homework Equations


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


The Attempt at a Solution


Would it be correct that: dt=(2m)(\frac{dv}{v}-\frac{du}{u})
If u and v are functions of t, then yes, you are correct.
 
SammyS said:
If u and v are functions of t, then yes, you are correct.

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
Airsteve0 said:
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))?

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