Derivative of Exponential Equation - Quick Question

[Airsteve0]

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: $dt=(2m)(\frac{dv}{v}-\frac{du}{u})$

 

[STEMucator]

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?

 

[Airsteve0]

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

 

[STEMucator]

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?

 

[Airsteve0]

v and u

 

[STEMucator]

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 $\frac{∂t}{∂u}$

 

[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) ?

 

[SammyS]

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.

 

[Airsteve0]

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))?

 

[STEMucator]

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.

 

[Airsteve0]

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