How did "v" come out of this derivative?

1. The problem statement, all variables and given/known data

3. The attempt at a solution

I basically understood how to attack the problem and my answer was very close to the key. The only thing bothering me is that v on the top. I cannot see how that could come up anywhere.

I know that the "r" is r = r(t), so that I am only taking the derivative of $$\ln(1 + w/r)$$. I don't see where the v actually is

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 Recognitions: Gold Member Homework Help The flux is expressed as a function of r. You're taking the derivative with respect to t. So, recall the chain rule of calculus: first take the derivative with respect to r and then multiply by the derivative of r with respect to t.

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How did "v" come out of this derivative?

 Quote by Dens Nope, implicit differentiation.

Here's the chain rule: If $y=f[u]$ and $u = g[x]$ are differentiable functions, then $\frac{dy}{dx}$=$\frac{dy}{du}$$\cdot$$\frac{du}{dx}$

You have the flux expressed as a function of r: $\Phi[r]$ and r is some function of time: $r[t]$.

So, the chain rule says $\frac{d\Phi}{dt}$ = $\frac{d\Phi}{dr}$$\cdot$$\frac{dr}{dt}$

This get's you the answer and it shows why the speed v appears.

How do you get the answer using implicit differentiation?

 Quote by TSny Here's the chain rule: If $y=f[u]$ and $u = g[x]$ are differentiable functions, then $\frac{dy}{dx}$=$\frac{dy}{du}$$\cdot$$\frac{du}{dx}$ You have the flux expressed as a function of r: $\Phi[r]$ and r is some function of time: $r[t]$. So, the chain rule says $\frac{d\Phi}{dt}$ = $\frac{d\Phi}{dr}$$\cdot$$\frac{dr}{dt}$ This get's you the answer and it shows why the speed v appears. How do you get the answer using implicit differentiation?
No you are right, I forget the $\phi$ on the LHS. I was thinking something like

Something = r as a function of t ==> differentiate both sides wrt t ==> implicitly differentiate RHS

http://www.wolframalpha.com/input/?i=D[ln%281+%2B+w%2Fr%28t%29%29%2Ct]